= Solution
For a <group action> of a <finite group> $G$ on a set $X$, let $G_x=\{g:gx=x\}$ be the <stabilizer subgroup> and $Gx=\{gx:g\in G\}$ the <orbit>. The <orbit-stabilizer theorem> is
$$
\boxed{|Gx|=[G:G_x],\qquad |G|=|Gx|\,|G_x|.}
$$
To prove it, the map from left <cosets> to the orbit, $gG_x\mapsto gx$, is well-defined because multiplying $g$ on the right by an element fixing $x$ does not change $gx$. It is surjective by the definition of the orbit. If $gx=hx$, then $h^{-1}g\in G_x$, so $gG_x=hG_x$; hence it is injective. The cardinality identity now follows from <Lagrange's theorem>.
Place the octahedron's vertices at $\pm e_1,\pm e_2,\pm e_3$. A symmetry fixes their centre, so it is an orthogonal map and must take an opposite pair of vertices to another opposite pair. Its matrix is thus a <signed permutation matrix>. Conversely every signed <permutation> of these coordinate directions preserves the vertices, edges and faces. Therefore the <full octahedral symmetry group> has
$$
\boxed{|G|=3!\,2^3=48.}
$$
These same signed <permutation> matrices preserve the cube with vertices $(\pm1,\pm1,\pm1)$, so the <group> is also the <symmetry group of a cube>. The opposite-vertex lines are the three coordinate axes, so \b[$|D|=3$]. The action on them is transitive, giving a line stabilizer of order $48/3=16$.
For the line $\mathbb Re_3$, the precise <axis stabilizer in the full octahedral symmetry group> consists of the block matrices
$$
\begin{pmatrix}A&0\\0&\varepsilon\end{pmatrix},\qquad
A\text{ a signed }2\times2\text{ permutation matrix},\quad\varepsilon\in\{1,-1\}.
$$
The eight possible $A$ are the symmetries of a square in the perpendicular plane, a <dihedral group> of order eight; the independent sign $\varepsilon$ reverses the selected axis. Thus
$$
\boxed{G_{\mathbb Re_3}\cong D_8\times C_2.}
$$
Here $D_8$ denotes the order-eight square-symmetry <group>. The selected line is unoriented: its two endpoints may be exchanged. Fixing an individual endpoint instead would give only the order-eight <subgroup> with $\varepsilon=1$.
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