 Solution

Burnside's lemma counts the orbits of a finite group action:Identify each face with its opposite vertex. Every symmetry induces the same permutation on these two four-element sets, so a permutation with $c$ cycles fixes $3^c{2}^c=6^c$ colorings. The rotational symmetry group is $A_4$, with one identity, eight three-cycles, and three products of two disjoint transpositions. Their cycle counts are four, two and two. Thus the rotation count isAllowing reflections gives the full tetrahedral group $S_4$. In addition there are six transpositions with three cycles and six four-cycles with one cycle. The count becomesEquivalently, 30 of the 141 rotational classes form 15 mirror pairs, while 111 rotational classes are their own mirror class.

The product of fixed-coloring counts is an instance of the orbit count for simultaneous colorings.