= Solution
Put the cube's vertices at $(\pm1,\pm1,\pm1)$. Squared distances between distinct vertices are $4,8,12$ according as one, two or three coordinates differ. Four mutually equidistant vertices cannot all differ in one coordinate, since the cube has no triangles, or in three coordinates, since each vertex has only one antipode. They must differ in two coordinates. The two <regular tetrahedra> are therefore precisely the sets $xyz=1$ and $xyz=-1$.
The full <symmetry group> $G$ consists of signed coordinate permutations, so $|G|=2^3\cdot3!=48$. These maps permute the two tetrahedra. The <stabilizer> $H$ of the first has order $24$ and acts faithfully on its four vertices: a linear map fixing the tetrahedron fixes a spanning set. Thus $H\cong S_4$. Central inversion $-I$ swaps the tetrahedra and commutes with every element of $H$, giving the <direct-product decomposition of the cube symmetry group>
$$
\boxed{G\cong S_4\times C_2.}
$$
Here $H$ preserves one tetrahedron; it is not the subgroup consisting only of rotations.
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