Use the Orbit Stabilizer Theorem to deduce the number of elements in the rotational symmetry group of the cube.
I can write $\operatorname{Stab}_G(v) = \left\{g \in G \mid g \cdot v = v\right\}$ and $\operatorname{Orb}_G(v) = \left\{g \cdot v \mid g \in G\right\}$
The orbit has size 8. Is it enough to say that it is 8 simply because there exists a symmetry such that a specific vertex can somehow get mapped to any of the others.
For the stabilizer, I considered a vertex on the top face of the cube. So I can see three rotations that would fix this vertex in place. Those are
- rotation about an axis going through this vertex and the vertex diagonally opposite and lower down,
- rotation about $2\pi$ and
- rotation about $-2\pi$.
Is this correct? So my answer would be $24 = (8 \times 3)$ by Orbit Stabilizer.
As a check, it might be helpful to note the rotational symmetry group of the cube (which we will call $G$) also acts on the set of 6 faces of the cube.
As anyone who has played with dice is aware, it is possible for any of the six faces to be "up", that is, there exists a rotational symmetry of the cube that maps a given face to any other (in a quaint display of harmony between math and ordinary English, we call the exhibiton of such a symmetry "rolling the die"). If one imagines one's cube with its center at $(0,0,0)$, with sides of length $2$, and one considers the face bounded by $(1,-1,1),(1,1,1),(1,1,-1)$ and $(1,-1,-1)$, it should be clear that the stabilizer of this face consists of the $4$ rotations (including the identity) about the $x$-axis in the $yz$-plane (which in fact is the rotational symmetry group of the SQUARE with vertices $(0,-1,1),(0,1,1),(0,1,-1),(0,-1,-1)$ in the $yz$-plane). The orbit-stabilizer theorem then tells us that:
$|G| = |\mathcal{O}_G(f)|*|\text{Stab}_G(f)| = 6*4 = 24$