 Solution

Place the cube's vertices at $(\pm 1,\pm 1,\pm 1)$. Every symmetry fixes its centre, and permutes the three perpendicular pairs of face normals, so it is a signed permutation matrix. Conversely every such matrix preserves the cube. The symmetry group of a cube acts transitively on its eight vertices: independent coordinate sign changes take $(1,1,1)$ to any vertex.

A symmetry fixing $(1,1,1)$ must permute its three incident edges. This gives an injective map from its stabilizer subgroup into $S_3$, since those three edge directions determine the linear transformation. All six permutations are realized by coordinate permutation matrices, which fix $(1,1,1)$. Thus the vertex stabilizer has order six. The orbit-stabilizer theorem now yieldsThe use of all symmetries includes reflections as well as the twenty-four elements of the rotational symmetry group of a cube.