 Solution

Take the cube with vertices $(\pm 1,\pm 1,\pm 1)$, and choose the edge $E=\{(1,1,z):-1\le z\le 1\}$. Here the symmetry group of a cube consists of all rigid symmetries, including reflections. A symmetry fixes the centre and permutes the three pairs of opposite faces, so its matrix is a signed permutation matrix.

A symmetry in the stabilizer subgroup of $E$ must preserve its midpoint $(1,1,0)$ and send its direction $(0,0,1)$ to itself or its negative. In this cube symmetry action on edges, its only choices are to interchange the first two coordinates and to reverse the third coordinate. Thus the stabilizer subgroup consists ofThe two generating reflections commute and have order two, so this subgroup is a Klein four-group, isomorphic to $C_2\times C_2$ and has order four. Permuting coordinates and changing signs carries $E$ to any edge. Its orbit of a group action is therefore the set of all twelve edges. The orbit-stabilizer theorem givesIf one restricts to orientation-preserving symmetries, only two of the four stabilizer elements remain, giving the rotational symmetry group of a cube of order $24$; the full symmetry group asked for here has order $48$.