I've shown that the symmetry group of a cube and a tetrahedron are both isomorphic to S4, but I am now trying to show that they are not conjugate when considered as subgroups of isometries of 3D space.
I cannot think of any kind of criteria that gives when two groups are not conjugate?
EDIT: I meant to say the rotations of a cube! So I've shown the full symmetry group of the tetrahedron is isomporphic to the rotations of a cube, however I need to show that they are not conjugate.
On
I think I've answered it now. If we just think of all the group elements as matrices. Then we want to show that these two groups are not conjugate. So we try and find an element to "conjugate with" from the isometry group. So we get $A = C^{-1}BC$ where $A$ is an element of the rotations of a cube, and B an element from the rotations and reflections of the tetrahedron.
It then follows that $det(A) = det(B)$ and since for rotations $det(A) = 1$, and for reflections $det(B) = -1$, then we see that certain elements cannot possibly be conjugate.
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
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.