Something's wrong with your first statement. If you consider orientation preserving 3D movements, the symmetry group of the cube is twice as big as that of the tetrahedron. And the same holds if you disregard orientation for both solids.
Note that taking $4$ out of $8$ vertices of a cube, you get a tetrahedron. Any symmetry of that tetrahedron induces a symmetry of the cube. However, the cube allows additional symmetries that map the tetrahedron to the "other" tetrahedron (the one with the other four vertices).
In other words, the fact that half the vertices of a cube form a tetrahedron shows that the tetrahedron group is a subgroup of index 2 in the cube group.
However, you do get $S_4$ if you count
- arbitrary isomometries of the tetrahedron (that is, just permutations of the four vertices) and
- orientation preserving isometries of the cube (that can be identified with permutations of the four spatial diagonals).
These two subgroups of the group of isometries cannot be conjugate to each other because the conjugate of an orientation preserving isometry is orientation preserving.