Is it a uniform polyhedron? Objective
Given a vertex figure consisting of regular convex polygons, determine whether it represents a convex uniform polyhedron.
What is a uniform polyhedron?
A uniform polyhedron is a polyhedron whose faces are regular polygons, while having the same vertex figure for each vertices. Generally a uniform polyhedron can be nonconvex, but only convex polyhedra will be considered in this challenge. (More precisely, the polyhedron is required to be vertex-transitive, but that's just another detail.)
What is a vertex figure?
In the context of a convex uniform polyhedron, a vertex figure is a list of the number of edges of polygons (in order) around a vertex. For example, a cube has vertex figure of (4.4.4).
Truthy inputs

(3.3.3) – Tetrahedron
(4.4.4) – Cube
(3.3.3.3) – Octahedron
(5.5.5) – Dodecahedron
(3.3.3.3.3) – Icosahedron
(4.4.N) for every N≥3 – N-gonal prism (It is a cube for N=4)
(3.3.3.N) for every N≥4 – N-gonal antiprism (It is an octahedron for N=3)
(3.6.6) – Truncated tetrahedron
(3.4.3.4) – Cuboctahedron
(3.8.8) – Truncated cube
(4.6.6) – Truncated octahedron
(3.4.4.4) – Rhombicuboctahedron
(4.6.8) – Truncated cuboctahedron
(3.3.3.3.4) – Snub cube
(3.5.3.5) – Icosidodecahedron
(3.10.10) – Truncated dodecahedron
(5.6.6) – Truncated icosahedron
(3.4.5.4) – Rhombicosidodecahedron
(4.6.10) – Truncated icosidodecahedron
(3.3.3.3.5) – Snub dodecahedron

Rotations and reversions (generally, all dihedral permutations) of these lists are also truthy. For example, (4.6.8), (4.8.6), (6.4.8), (6.8.4), (8.4.6), (8.6.4) are all truthy.
Falsy examples

(3.3.3.3.3.3) – Triangular tiling; not a polyhedron.
(5.5.5.5) – Order-4 pentagonal (hyperbolic) tiling; not a polyhedron.
(3.3.4.4) – Cannot be uniform. Note that this is different from (3.4.3.4).

Don't care situations
An input is expected to have at least 3 entries, and to consist of integers that are at least 3. Otherwise, the challenge falls in don't care situation.

(5/2.5/2.5/2) – Great stellated dodecahedron; not convex.
(3.3) – Triangular dihedron; not Euclidean.
(2.2.2) – Triangular hosohedron; not Euclidean.
(3/2.3/2.3/2) – Retrograde tetrahedron.
(1)
(-3)
()
Python 3, 190 bytes

def f(F):s=&quot;&quot;.join(hex(k)[2]for k in F);F[1:]in[[4,4],[3]*3]or{s,s[::-1]}&amp;{*&quot;555 333 366 388 3aa 466 566 468 46a 3434 3444 3454 3535 33333 33334 33335&quot;.split()}and max(F)&lt;16or f(F[1:]+F[:1])

Try it online!
Takes input as a list of integers representing the vertex figure. The function errors (RecursionError) if the vertex figure is not a uniform polyhedron, otherwise there is no error.
I tried several schemes of organizing the finite classes into a smart way that takes advantage of patterns, but hardcoding all possibilities reigned superior since it is a relatively small set.
Explanation
def f(F):
    # F is a rotation of the input vertex figure; initially is the input vertex figure
    # Convert to string for easier comparison later in the code
    s=&quot;&quot;.join(hex(k)[2]for k in F)
    # Test true if the permutation is N.4.4.4 or N.3.3.3
    (F[1:]in[[4,4],[3]*3]or
    # Test truthy if permutation (or its reverse) is in
    #   3.3.3, 3.6.6, 3.8.8, 3.10.10, 4.6.6, 5.6.6, 4.6.8, 4.6.10,
    #   3.3.3.N, 3.4.3.4, 3.4.4.4, 3.4.5.4, 3.5.3.5
    #   3.3.3.3.3, 3.3.3.3.4, 3.3.3.3.5
    {s,s[::-1]}&amp;{*&quot;555 333 366 388 3aa 466 566 468 46a 3434 3444 3454 3535 33333 33334 33335&quot;.split()}
    # Numbers greater than 15 would convert into the most-significant hexit when converted to a string,
    # causing 0x43 to match the same as 0x4,
    # so we need to check that none of this happened if we want a bugfree string search
    and max(F)&lt;16
    # If we tested truthy, then terminate
    # Otherwise, recurse with the vertex figure cyclically rotated left one
    or f(F[1:]+F[:1]))