= Solution
Each nonidentity element of the <special orthogonal group> $\operatorname{SO}(3)$ 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 $g\ne1$ and $h\in G$, then $hv$ is fixed by $hgh^{-1}\ne1$. 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 $n_1,\ldots,n_r$, each at least two. Count pairs $(g,v)$ with $g\ne1$ and $gv=v$. Counting by $g$ gives $2(N-1)$. Counting by <group orbits> and using the <orbit-stabilizer theorem> gives
$$
\sum_{j=1}^r\frac N{n_j}(n_j-1)=2(N-1),\qquad
\sum_{j=1}^r\left(1-\frac1{n_j}\right)=2-\frac2N<2.
$$
Every summand is at least $1/2$, so $r<4$: \b[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
$$
\boxed{12,\quad20,\quad30.}
$$
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