 Solution

Each nonidentity element of the special orthogonal group $\mathrm{SO}(3)$ is a rotation fixing exactly the two unit vectors along its axis. Since $G$ is a finite group, $\Omega$ is finite. If $v\in \Omega$ is fixed by $g\ne 1$ and $h\in G$, then $hv$ is fixed by $hg{h}^{-1}\ne 1$. Hence $G$ acts on $\Omega$ by a group action.

Let $N=|G|>1$, let $r$ be the number of orbits of a group action, and choose representatives with stabilizer subgroups of orders $n_1,\dots ,n_r$, each at least two. Count pairs $(g,v)$ with $g\ne 1$ and $gv=v$. Counting by $g$ gives $2(N-1)$. Counting by group orbits and using the orbit-stabilizer theorem givesEvery summand is at least $1/2$, so $r<4$: there are at most three orbits. For the trivial group, $\Omega$ is empty and the assertion also holds.

For the rotation group of a regular dodecahedron, $N=60$. The three group orbits are the unit vectors towards the 12 face centres, 20 vertices and 30 edge midpoints, whose stabilizer subgroup orders are respectively $5,3,2$. Rotations about these axes exhaust the nonidentity rotations. Consequently the orbit sizes are