Question:

Using Burnside's lemma, count colorings fixed by each symmetry of the cube. There are $2^8 = 256$ total colorings when vertices are fixed. Under 90° rotation about face midpoints: $2^4 = 4$ fixed colorings. Under 120° rotation about opposite corners: $2^4 = 16$ fixed colorings (splits 8 vertices as $1+3+3+1=8$). Under 180° rotation about face midpoints: $2^4 = 16$ fixed colorings (splits as $2+2+2+2=8$). Under 180° rotation about opposite edge midpoints: $2^4 = 16$ fixed colorings (splits as $2+2+2+2=8$). The number of orbits is $\frac{1}{24}(1 \times 256 + 6 \times 4 + 8 \times 16 + 3 \times 16 + 6 \times 16) = 23$.