User blog comment:Edwin Shade/The Nano-Graham Problem/@comment-1605058-20171008195816

There is a standard method of solving such problems, which proceeds using. Let me explain how to use it to solve your problem.

We consider all possible symmetries of the cube. In total, there are 48 of them, and you can find their list here. For each symmetry, we have to count how many colorings there are which are invariant under this symmetry, i.e. we get the same coloring after applying the symmetry. Here it goes:

Trivial symmetry: Every coloring is invariant, so we get \(2^{12}\).

90° rotations around a line through midpoints of faces (6 of them): Say we take a rotation fixing top and bottom face. A coloring is invariant iff top face's edges have the same color, bottom face's edges have the same color and side edges have the same color. This gives \(2^3\) colorings.

180° rotations around a line through midpoints of faces (3 of them): We get 6 pairs of edges which have to be colored the same, so \(2^6\) colorings.

180° rotations around a line through midpoints of edges (6 of them): There are 5 pairs of edges which have to be colored the same, plus the two edges through which the rotation line goes. This gives \(2^7\) colorings.

120° rotations around a line through vertices (8 of them): We get 4 triples, \(2^4\) colorings.

Central symmetry: 6 pairs, \(2^6\) colorings.

Reflection in a plane through four edge centers (3 of them): 4 pairs, 4 single edges, \(2^8\) colorings.

Reflection in a plane containing two edges (6 of them): 5 pairs, 2 single edges, \(2^7\) colorings.

60° "Rotoreflections" (8 of them): (a picture) Two sextuples of edges, \(2^2\) colorings.

90° "Rotoreflections" (6 of them): (a picture) Three quadruples of edges, \(2^3\) colorings.

Burnside's lemma now says that the number of colorings with symmetric colorings counted once is (the sum over all symmetries of the number of invariant colorings)/(the number of symmetries), i.e. \[(1\cdot 2^{12}+6\cdot 2^3+3\cdot 2^6+6\cdot 2^7+8\cdot 2^4+1\cdot 2^6+3\cdot 2^8+6\cdot 2^7+8\cdot 2^2+6\cdot 2^3)/48=144.\]