Let $T$, resp. $I$, denote the group of rotational symmetries of tetrahedreon, resp. icosahedron.
Fix a face of each of these polyhedrons; let centers of these polyhedrons be $(0,0,0)$ in $\mathbb{R}^3$.
For any of these groups $T$ and $I$, let $\rho$ be the rotation of smallest positive degree, clockwise, around the axis passing through center of a face and origin (center of body).
Q. In the group $T$, $\rho,\rho^{-1}$ are not conjugate but in group $I$, $\rho,\rho^{-1}$ are conjugate; why?
Two possible answers are there:
(1) (A long jump) Icosahedron has antipodal symmetry $(v\mapsto -v)$ but tetrahedron has not.
(2) $T$ is isomorphic to $A_4$ in which $(123)$ and $(132)$ are not conjugate but $I$ is $A_5$ in which $(12345)$ and its inverse $(15432)$ are conjugate.
But my problem is the following: these groups $T,I$ are visualizable, then can't we visualize the conjugacy or non-conjugacy of the objects under consideration? How can we answer Q. pictorially?
If you write "spin the icosahedron upside down, then give it a twist by $36^\circ$" as $r$, then $r\rho r^{-1}=\rho^{-1}$. This is because a rotation viewed from above as "clockwise", if viewed upside-down (geometrically, conjugating by $r$ is "rotating yourself by $r$"), becomes "counterclockwise".
Similarly, a cube can be spun upside down (and there's no need to add a little twist this time!), so $\rho$ is conjugate with $\rho^{-1}$ in the rotation group of cube/octahedron.
Tetrahedron does not have opposite faces, so it can't be spun upside down.