= Solution
<Burnside's lemma> counts the orbits of a finite <group action>:
$$
|\Omega/G|=\frac1{|G|}\sum_{g\in G}|\operatorname{Fix}(g)|.
$$
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 $3^c2^c=6^c$ 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 is
$$
\boxed{\frac{6^4+8\cdot6^2+3\cdot6^2}{12}=141}.
$$
Allowing 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 becomes
$$
\boxed{\frac{6^4+6\cdot6^3+3\cdot6^2+8\cdot6^2+6\cdot6}{24}=126}.
$$
Equivalently, 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>.
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