= Solution
Use the right-action convention printed in the paper: $\omega1=\omega$ and $(\omega g)h=\omega(gh)$. The <orbit of a group action> and <stabilizer> are
$$
\operatorname{Orb}(\omega)=\{\omega g:g\in G\},\qquad H=\operatorname{Stab}(\omega)=\{g\in G:\omega g=\omega\}.
$$
The 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 $k\in G$,
$$
k\in\operatorname{Stab}(\omega g)\iff\omega gkg^{-1}=\omega\iff gkg^{-1}\in H,
$$
proving \b[$\operatorname{Stab}(\omega g)=g^{-1}Hg$].
The rotational action on the eight vertices of a <cube> is \b[<transitive>]. Model the vertices as $(\pm1,\pm1,\pm1)$. 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 \b[the <stabilizer> is cyclic of order three].
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