 Solution

Use the right-action convention printed in the paper: $\omega 1=\omega$ and $(\omega g)h=\omega (gh)$. The orbit of a group action and stabilizer areThe identity belongs to $H$, products of elements of $H$ fix $\omega$, and $\omega g=\omega$ implies $\omega g^{-1}=\omega$. Thus $H$ is a subgroup. For $k\in G$,proving $\mathrm{Stab}(\omega g)=g^{-1}Hg$.

The rotational action on the eight vertices of a cube is transitive. Model the vertices as $(\pm 1,\pm 1,\pm 1)$. For any desired sign triple, choose a permutation whose parity makes the corresponding signed permutation matrix have determinant one; it maps $(1,1,1)$ to that vertex and preserves the cube. A rotation fixing a vertex fixes the axis through that vertex, the centre and the opposite vertex. It cyclically permutes the three incident edges, giving precisely rotations by $0,2\pi /3,4\pi /3$. Thus the stabilizer is cyclic of order three.