What are the symmetries of the tetrahedron? Suppose I like combinatorics, and want to count how many ways to paint the faces of a tetrahedron using a pallet of $x$ colors. 
I don't want to over count cases where I could just rotate one painted tetrahedron to look like another. So my idea is that you could find all the cycle decompositions of the elements in the group of symmetries of the tetrahedron, paint each cycle any of the $x$ colors, add them all up, and divide by the number of elements in the group, to count the distinct ways to paint, correct?
So my question boils down as, what is the group of symmetries of the tetrahedron? I would be happy just to know the elements in terms of rotations and flips around axes, and could probably figure out the cycle decompositions from there.
 A: I think you're overcomplicating things. A coloured (regular) tetrahedron is chiral if and only if all four faces have different colours. Thus each of the $\binom x4$ combinations of $4$ different colours leads to $2$ distinguishable colourings, and the $\binom xk$ combinations of $k<4$ colours each lead to $\binom3{k-1}$ distinguishable colourings (one for each way of partitioning $4$ onto the $k$ colours), so the total number is
$$2\binom x4+3\binom x3+3\binom x2+x\;.$$
A: Do reflection symmetries count for you, or are you restricting your interest to rotations?
If reflections are allowed, then you can realize any permutation of the faces you want, thus the symmetry group is the symmetric group $S_4$. Otherwise, half of these permutations are forbidden, and the symmetry group is the alternating group $A_4$.
In terms of painting sides, what this means is that any set of 4 different colors can be used in exactly two ways (one if you allow reflections). Sets where at least one color appears more than once can only be used in one way.
A: As this post asks for the group $G$ of symmetries of the faces of the tetrahedron for use with the Polya Enumeration Theorem (PET) and this group is quite small we can do the calculation by hand and compute the cycle index.
We enumerate the elements of $G$ in turn and list their contributions to the cycle index, where we consider rigid motions i.e. no reflections. There is the identity, which contributes $$a_1^4.$$ There are two rotations about an axis passing through the center of a face and the opposite vertex and there are four such axes, giving the contribution $$4\times 2 a_1 a_3^.$$ Finally there are three rotations about the centers of two opposite edges by 180 degrees, giving $$3\times a_2^2.$$
This gives for the cycle index that
$$Z(G) = \frac{1}{12}\left(a_1^4 + 8 a_1 a_3 + 3 a_2^2\right).$$
Now with $X$ colors we get that
$$Z(G)(C_1+C_2+\cdots+C_X)_{C_1=1, C_2=1, \ldots C_X=1}
= \frac{1}{12} (X^4+ 8 X^2 + 3 X^2) =
 \frac{1}{12} (X^4+ 11 X^2).$$ 
This matches the answer by Joriki and produces the sequence
$$1, 5, 15, 36, 75, 141, 245, 400, 621, 925,\ldots$$
and one might be suprised to find that this sequence is sufficiently important to appear as
OEIS A006008.
Here are some more PET computations at this MSE link.
