 Solution

Centre the cube at the origin with faces having normals $\pm e_1,\pm e_2,\pm e_3$. Every rotational symmetry fixes the centre and permutes these normals, so its rotation matrix is a signed permutation matrix of determinant one. Conversely any such matrix permutes the cube’s coordinates up to signs and preserves the cube. For each of $3!$ coordinate permutations, four of the eight choices of signs give determinant one. Consequently the rotational symmetry group of a cube hasThis counts the rotations directly, without an unverified stabilizer calculation. The TeX’s $2^k$ is damaged transcription: the PDF prints 24.

Let $X$ be the $3^6$ face colourings. A colouring fixed by a rotation must be constant on each permutation cycle of faces, giving $3^c$ fixed colourings if the rotation has $c$ face cycles. The rotation types are:

There are three face axes, each with two quarter-turns and one half-turn; six edge axes, each with one half-turn; and four vertex axes, each with two nonidentity third-turns. These rotations are distinct and, with the identity, total 24, so the table is exhaustive. Face-axis rotations fix their two axial faces; the remaining four faces cycle or pair. Edge half-turns pair all faces, while vertex third-turns form two triples.

Burnside lemma now givesThe three colours are labelled, orthogonal reflections are excluded, and no condition requires every colour to appear.