= Solution
Choose $Q\in SO(n+1)$ with $Qe_{n+1}=v$. Conjugation by $Q$ identifies
$$
S_v=\{R\in SO(n+1):Rv=v\}
$$
with
$$
\left\{
\begin{pmatrix}A&0\\0&1\end{pmatrix}:A\in SO(n)
\right\}.
$$
It follows that the <unit-vector stabilizer in a special orthogonal group> is an embedded submanifold diffeomorphic to $SO(n)$, and
$$
\boxed{\dim S_v=\frac{n(n-1)}2}.
$$
At $R\in S_v$, its tangent space is
$$
T_RS_v=\{RA:A^T=-A,\ Av=0\}.
$$
For any $v\ne w$, both $S_v$ and $S_w$ contain the identity. At the identity,
$$
T_IS_v=\{A\in\mathfrak{so}(n+1):Av=0\},
\qquad
T_IS_w=\{A\in\mathfrak{so}(n+1):Aw=0\}.
$$
Identify $\mathfrak{so}(n+1)$ with the exterior square $\Lambda^2\mathbb R^{n+1}$. The orthogonal complement of $T_IS_v=\Lambda^2(v^\perp)$ is $v\wedge v^\perp$. If $v$ and $w$ are linearly independent, the nonzero bivector $v\wedge w$ belongs to the orthogonal complements of both tangent spaces. Their sum is therefore a proper subspace of $T_ISO(n+1)$, so the intersection is not <transverse intersection>[transverse]. If $w=-v$, then $S_w=S_v$, whose tangent space is proper, so they are again not transverse. Consequently
$$
\boxed{S_v\text{ and }S_w\text{ are never transverse when }v\ne w},
$$
as summarized by <distinct unit-vector stabilizers are not transverse>.
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