Some questions on building permutations from boolean functions I've seen multiple examples of boolean functions being used as a permutation.
For example the Keccak Chi : 2.3.1 function:

from https://keccak.team/figures.html
Or as a formula: for $i=\{0..4\}$ $A_i=a_i \oplus (\neg a_{i+1} \wedge a_{i+2})$ with indexes calculated modulo 5
The first question would be, what is the rationale (or proof) why this is a permutation?
The second, related one: What are the properties, the boolean function has to satisfy that it results in a permutation?

And now regarding the inverse of such a permutation.
Are there any generic methods/algorithms to find the inverse of such a construction?
Also, what are the major contributing factors for the complexity of the inverse (number of variables, algebraic degree etc.)?
And if such a method is applied to a larger input - say $i=\{0..127\}$, is the inverse more difficult to calculate, if the function has only a few (like the 3 for Chi) or many, say 128, input variables?
Any answers/pointers are appreciated.
The general algebraic question is multifaceted and can be quite complicated. Some depend on vector space, some on extension field properties.
As mentioned in the comments checking the property can be simpler.
I answered a related question Examples of multi output bit balanced Boolean functions
Nyberg’s articles mentioned there are
K. Nyberg, Differentially uniform mappings for cryptography, 1993 and
K Nyberg, Perfect nonlinear S-boxes, 1992
both easily locatable on google scholar.
Edit: The keccak $\chi$ maps $\{0,1\}^5$ to itself.
I will use $a_i$ as input and $A_i$ as output variables as in the edited question.
Counting indices modulo 5, if there is no $i$ such that $(a_i,a_{i+2})=(0,1)$ then $\chi$ has a fixed point for that input. Let $W=\{i: (a_i,a_{i+2})=(0,1)\},$ then the general mapping just inverts the bits belonging to $i.$
Note that the sets $J_i,J_j$ where $J_i=\{i,i+2\}$ are disjoint except when $j=i+2$ or $i=j+2.$ So there is no ambiguity for determining the inverse unless we are in this special case, thus the inverse exists except in this special case. But even in this case the patterns
$(a_i,a_{i+2},a_{i+4})$ which result in bitflips are unambiguous.
If $(a_i,a_{i+2},a_{i+4})=(1,0,0)$ then $a_{i+1}$ will be flipped but not $a_{i+3}$. So $A_{i+1}=1\oplus a_{i+1},$ and $A_{i+3}=a_{i+3}.$
If $(a_i,a_{i+2},a_{i+4})=(1,0,1)$ then $a_{i+1}$ will be flipped but not necessarily $a_{i+3}$, that will depend on the value of $a_{i+6}=a_{i+1}$. But that bit is not impacted by the previous argument since $J_i$ and $J_j$ are disjoint if $i=j+1\pmod 2.$
So a unique inverse mapping exists.
Remark: In general going between &quot;basis independent&quot; extension field formulations of permutations vs &quot;basis dependent&quot; bit vector permutations is hardly straightforward. I don't see an immediate basis independent extension field formulation for this permutation, and as pointed out in the comments to the question such formulations obtained (say) by Lagrange interpolation, can be quite complicated and high degree.