 Solution

Choose $Q\in SO(n+1)$ with $Q{e}_{n+1}=v$. Conjugation by $Q$ identifieswithIt follows that the unit-vector stabilizer in a special orthogonal group is an embedded submanifold diffeomorphic to $SO(n)$, andAt $R\in S_v$, its tangent space is

For any $v\ne w$, both $S_v$ and $S_w$ contain the identity. At the identity,Identify $\mathfrak{so}(n+1)$ with the exterior square ${\Lambda }^2{\mathbb{R}}^{n+1}$. The orthogonal complement of $T_I{S}_v={\Lambda }^2(v^{\perp })$ 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 $T_{I}SO(n+1)$, so the intersection is not transverse. If $w=-v$, then $S_w=S_v$, whose tangent space is proper, so they are again not transverse. Consequentlyas summarized by distinct unit-vector stabilizers are not transverse.