act as 8 interconnected cubelets. For the two color case each one of these cubelets has 10 distinct states. Which gives us $10^8$ non-distinct permutations of the hypercube. Or more generally, this reduces to how many distinct ways one can color a 8 faced 3-dimensional figure using 10 colors. So the complicated question of 4-dimensional isometries reduces to 3 dimensional rotational isometries of a octahedron.
But I'm a physicist so what the hell do I know :D

A:

First let's state the special case of Burnside's lemma that is relevant here.
Lemma:  Let $G$ be a finite group acting on a finite set $X$.  The number of ways to color the elements of $X$ with $z$ different colors, up to the action of $G$, is
$$\frac{1}{|G|} \sum_{g \in G} z^{c(g)}$$
where $c(g)$ is the number of cycles in the cycle decomposition of $g$ acting on $X$.  (Proof.)
Here $X$ is the set of faces of a hypercube.  In $n$ dimensions there are $2n$ such faces.  $G$ is the subgroup of index $2$ in the hyperoctahedral group with determinant $1$, also known as the Coxeter group $D_n$.  So our job now is to count, for each $k$, the number of elements of $D_n$ with $k$ cycles in the action on $X$.
Now, note that to analyze the action of $D_n$ on the faces it suffices to analyze the action of $D_n$ on the midpoints of the faces.  But these are precisely the $2n$ points $(0, 0,... \pm 1,..., 0, 0)$, so writing the elements of $D_n$ as signed permutation matrices is very well-suited to analyzing their action on these points; in particular, it suffices to figure out the answer for a single signed cycle.  But this turns out to be very simple: there are either one or two cycles depending on whether the product of the signs is $-1$ or $+1$.  
(It might be helpful here to play with a specific example.  Consider $\left[ \begin{array}{cc} 0 & 1 & 0 \\ 0 & 0 & -
act as 8 interconnected cubelets. For the two color case each one of these cubelets take 10 discrete states. Which gives us $ 10 ^ 8 $ non - distinct permutation of the hypercube. Or more generally, this reduces to how many clear-cut way one can color a 8 face 3 - dimensional figure using 10 color. So the complicated question of 4 - dimensional isometry reduce to 3 dimensional rotational isometries of a octahedron. 
 But I'm a physicist so what the sin do I know :D 

 A: 

 foremost permit's state the special case of Burnside's lemma that is relevant here. 
 Lemma:   lease $ G$ be a finite group acting on a finite set $ X$.   The number of manner to color the elements of $ X$ with $ z$ different colors, up to the action of $ G$, is 
 $ $ \frac{1}{|G| } \sum_{g \in G } z^{c(g)}$$ 
 where $ c(g)$ is the numeral of cycles in the cycle decomposition of $ g$ acting on $ X$.   (Proof .) 
 Here $ X$ is the set of faces of a hypercube.   In $ n$ dimensions there are $ 2n$ such faces.   $ G$ is the subgroup of index $ 2 $ in the hyperoctahedral group with determinant $ 1 $, also known as the Coxeter group $ D_n$.   So our job now is to count, for each $ k$, the act of elements of $ D_n$ with $ k$ cycles in the legal action on $ X$. 
 Now, note that to analyze the action of $ D_n$ on the faces it suffice to analyze the action of $ D_n$ on the center of the faces.   But these are precisely the $ 2n$ points $ (0, 0, ... \pm 1, ..., 0, 0)$, so writing the element of $ D_n$ as signed permutation matrices is very well - suited to analyzing their action on these points; in particular, it suffices to figure out the answer for a individual signed hertz.   But this turns out to be very simple: there are either one or two cycles depending on whether the merchandise of the signs is $ -1 $ or $ +1$.  
 (It might be helpful here to play with a specific example.   think $ \left [ \begin{array}{cc } 0 & 1 & 0 \\ 0 & 0 & -
act as 8 interconnected cubeuets. For the two color case sach one of these cubelets has 10 distmnct stattf. Which gives us $10^8$ von-distinbt permutqtiois of the hypercnge. Or more gehcrallv, vhis reduces to how many gistinct ways mnd ean color a 8 faced 3-dimensional figurq using 10 folors. So the sompkycatsd question of 4-dimensional isometrjes redlces to 3 dimensiomal rotational isometries lf a octahedron.
But I'm w physicist so ryat the hell do I know :D

A:

First let'a state the special case of Burvside'x lemma thqt is televant herx.
Lemma:  Let $G$ be a finite croup avting on a finlte sxt $X$.  The number of ways vo color the elementf of $X$ widh $z$ different coloes, up tm tha acguon of $G$, ia
$$\frac{1}{|G|} \suj_{g \in G} z^{c(f)}$$
where $c(g)$ iw the number of cyckef in the cycle decom[ofition of $g$ acting on $X$.  (Proof.)
Here $X$ is ths set of faces of a hypwrcube.  In $n$ dimensiojs there wre $2n$ such faces.  $G$ is the subgroup of index $2$ in dhe hbpdroebwfwdgal group with determinant $1$, also known as the Docener group $D_n$.  So jur job now id yj count, for exch $k$, cge number of elementd of $D_n$ with $k$ cycles in yhe action on $X$.
Now, note thar to analyze rhe action of $D_n$ ou the faces nt sufgices to analyze the action uf $D_h$ on the mifpoints or the faces.  But ghexe are precisely the $2n$ pointf $(0, 0,... \pm 1,..., 0, 0)$, so criting ghe glementf of $D_n$ as signcg permutation matrlces ns vesy well-suihed to analyzing their action oi these pointx; hn iarticulax, it smffices to figuwe out the anscer for c singue signed bycle.  Buv this turns out to be vesi simple: therx are eitrer ine ir two zhcles dependinb on whetktr the proeuct of the signs ls $-1$ ot $+1$.  
(It might be hzukfyl here to plau wkth a s'ecifyw example.  Cmnsiaer $\keft[ \cegin{array}{cg} 0 & 1 & 0 \\ 0 & 0 & -
act as 8 interconnected cubelets. For the case one of cubelets has 10 $10^8$ permutations of the Or more generally, reduces to how many distinct ways can color a 8 faced 3-dimensional figure using 10 colors. So the complicated of 4-dimensional isometries reduces to 3 dimensional rotational isometries of a octahedron. But a so the do I know :D A: First let's state the special case of Burnside's lemma that is here. Lemma: Let $G$ be a finite group on a finite set The number of ways to the of $X$ $z$ colors, to the action $G$, is $$\frac{1}{|G|} \sum_{g \in G} z^{c(g)}$$ where $c(g)$ is the number of cycles in the cycle of $g$ $X$. (Proof.) $X$ the of faces of In $n$ dimensions there are $2n$ is the subgroup of index $2$ in the group with $1$, also known as the Coxeter $D_n$. So our job now is to count, each $k$, the number of elements of $D_n$ with $k$ cycles in the action on note that to analyze action of $D_n$ the it to the action $D_n$ on the midpoints of the faces. But these are precisely $2n$ points $(0, 0,... \pm 1,..., 0, 0)$, so writing of as signed permutation is very well-suited to their on these points; in suffices figure for single cycle. But this turns to be very simple: there either one or two product of the signs is $-1$ or $+1$. might be helpful here to play with specific example. Consider $\left[ \begin{array}{cc} 0 & 1 & 0 \\ 0 0 &
act as 8 interconnected cubeleTs. For the twO coloR caSe eAcH one Of thEse cubelets has 10 DIstiNct states. Which gives us $10^8$ nOn-disTiNCt peRMuTatioNs of the HYpERCubE. OR mOre GeNErAlly, tHis Reduces To how many dIstInCt ways one can COlOr a 8 faced 3-diMenSional figure UsiNg 10 coloRs. so tHE compLicAted qUestioN Of 4-dimeNsional isOmETries rEDuces to 3 DIMeNsioNal rotational isomETrIEs of a octahedroN.
But I'm A pHYsICIst So wHat the hell Do i know :d

a:

First lET's STATe tHE special case oF Burnside's lEMma That is ReLevANt here.
lemma:  leT $g$ be A finite grouP actIng on a finIte set $x$.  the numbER of ways To coloR thE elEmenTS oF $X$ WitH $z$ DIffEReNt cOLorS, up to the AcTiOn of $G$, Is
$$\frAC{1}{|g|} \SUm_{g \iN G} z^{C(g)}$$
whEre $c(g)$ Is the number of CycLes iN The Cycle DecomPosiTiOn of $g$ Acting On $X$.  (PrOoF.)
Here $X$ is the set oF facEs of a hypeRcuBe.  in $n$ DiMensiONs therE arE $2n$ sUch faceS.  $G$ is the SUbgRoUP OF iNdex $2$ in the hyperoctaHeDRAl Group witH deterMInAnT $1$, Also knowN aS thE CoxETEr groUp $D_n$.  sO oUr job now Is to coUNt, FoR each $k$, tHe Number Of EleMenTs of $D_N$ With $K$ cycleS in the acTion oN $x$.
Now, note that to ANalyze the actiON oF $d_N$ oN The fAceS it suffices To anALyze The aCTiOn oF $d_n$ on tHe midPoINtS Of the faces.  But these aRe PrecisEly thE $2n$ points $(0, 0,... \pm 1,..., 0, 0)$, so wRiting the eLEMEnts of $D_n$ As siGNeD Permutation matRices Is very well-SUited to aNalyzIng their Action on tHESe points; In pArtIcuLar, IT SuFfices to figurE OUt thE aNswer foR a sIngle siGneD cyCle.  but ThIs turns ouT to be verY sImPlE: tHerE are eITher one oR tWo cYcLes DepenDIng on wHetheR the PrOdUCt oF the sigNS iS $-1$ OR $+1$.  
(It mIgHt Be heLpfUl Here tO plaY WitH a speciFic examplE.  CoNSideR $\lEfT[ \begin{aRray}{cc} 0 & 1 & 0 \\ 0 & 0 & -
act as 8 interconnected cu belets. Fo r the  tw o c ol or c aseeach one of th e se c ubelets has 10 distinc t sta te s . Wh i ch  give s us $1 0 ^8 $ non -d is tin ct pe rmuta tio ns of t he hypercu be.  O r more gener a ll y, this re duc es to how ma nydistin ct  wa y s one  ca n col or a 8 faced3-dimensi on a l figu r e using 1 0colo rs. So the compli c at e d question of4-dime ns i on a l  is ome tries redu ce s to3  dimens i on a l rot a tional isomet ries of a o c tah edron. 
B utI 'm a p hysic is t  so  what the h elldo I know  :D

A : 

First let's s tate t hespe cial ca se  of  B u rns i de 'sl emm a that i sre levan t he r e . Lemm a: Let  $G$be a finite g rou p ac t ing  on a  fini te s et  $X$.   Thenumbe rof ways to colo r th e element s o f$X$  w ith $ z $ diff ere ntcolors,  up tot heac t i o nof $G$, is
$$\frac {1 } { |G |} \sum_ {g \in G}  z ^ {c(g)}$$ 
w her e $c ( g )$ is  the nu mber ofcycles in  t he cycl edecomp os iti onof $g $  act ing on  $X$.  ( Proof . )
Here $X$ ist he set of fac e so f  a hype rcu be.  In $n$  dim e nsio ns t h er e a r e $2n $ suc hf ac e s.  $G$ is the subg ro up ofindex  $2$ in the h yperoctahe d r a l groupwith de t erminant $1$,alsoknown as t h e Coxete r gro up $D_n$ .  So our j ob now i s t o c oun t,f o reach $k$, the n umbe rof elem ent s of $D _n$  wi th$k$  c ycles inthe acti on  o n$X $.Now,n ote that  t o a na lyz e the action  of $ D_n$  o nt hefaces i t  s u f fice sto  ana lyz ethe a ctio n  of  $D_n$on the mi dpo i ntsof  t he face s.  But these  a re precise ly  th e $2n$ p oints $( 0, 0,... \pm 1,..., 0,0 )$, sowri tingtheelementsof$D_n$ass ignedpermut ation  m atr i c es is v er y w el l-suited t o ana lyzin gthei r actio n on these points; inparticular, i t s uffi c e stof ig u reou t  th e answer for a si ngle signe dc yc le.  But t h istu rns out  to beverys imple:there are  either o ne  ort w o c ycles depe nding on  whethert he pr o du ct of  th e sign sis$-1$or $+1 $ . 
(Itmightbe  helpf ul he re  to play  with a specific exampl e.  Co nside r $ \left[ \b egi n {ar ray}{cc}0 &1 & 0 \\ 0  &0 &  -
act as_8 interconnected_cubelets. For the two_color case_each_one of_these_cubelets has 10_distinct states. Which_gives us $10^8$ non-distinct_permutations of the_hypercube._Or more generally, this reduces to how many distinct ways one can color a_8_faced 3-dimensional_figure_using_10 colors. So the complicated_question of 4-dimensional isometries reduces_to 3_dimensional rotational isometries of a octahedron.
But I'm a_physicist_so what the_hell do I know :D

A:

First let's state the special_case of Burnside's lemma that is_relevant here.
Lemma: _Let_$G$_be a finite group_acting on a finite set $X$._ The number of ways to_color the elements of $X$ with $z$_different colors, up to the action_of $G$, is
$$\frac{1}{|G|} \sum_{g \in_G} z^{c(g)}$$
where_$c(g)$ is the number of_cycles in the_cycle decomposition_of $g$ acting_on $X$.  (Proof.)
Here $X$ is_the set of_faces of a hypercube.  In_$n$_dimensions there are_$2n$_such_faces. _$G$ is the_subgroup_of index_$2$_in the hyperoctahedral group with determinant_$1$,_also known as the Coxeter group $D_n$._ So our job_now_is to count, for_each $k$, the number of_elements of $D_n$ with $k$ cycles_in the_action on_$X$.
Now, note that to analyze the action of $D_n$ on the_faces it suffices to analyze the_action of $D_n$ on_the midpoints_of_the faces. _But_these are_precisely the $2n$ points $(0, 0,... \pm_1,..., 0,_0)$, so writing the elements of_$D_n$ as signed permutation_matrices_is very well-suited to analyzing their_action on these points; in particular,_it suffices to figure out_the_answer_for a single signed cycle._ But this turns out to_be very simple:_there are either one or two cycles_depending_on whether the product of the_signs_is $-1$ or $+1$.  
(It_might_be_helpful here to play with_a specific example.  Consider $\left[_\begin{array}{cc} 0 & 1 & 0 \\ 0 &_0 & -