# Conjugating rotation by another rotation

If $g ∈ \mathrm{SO}(3)$ is the rotation about axis $p$ by angle $α$, and $h$ is a rotation mapping $p$ to another line $q$, then $g$ conjugated by $h$ is the rotation about $q$ by the same angle $α$. I've seen this fact presented as obvious, but I don't find 3D rotations trivial to visualize in general, and would like to see a formal proof. This could presumably be brute-forced in a number of ways, but I'm interested in an argument that avoids lengthy computation.

Since $hgh^{-1}$ obviously fixes $q$, that is indeed the axis of the resulting rotation, so it would suffice to prove it rotates $q^{\bot}$ by $α$.

An easy way to see this is to use Rodrigues's rotation formula: $$g{\bf v}=\cos \alpha \,{\bf v}+(1-\cos \alpha)({\bf u}\cdot {\bf v}){\bf u}+\sin \alpha \,{\bf u}\times {\bf v}$$
where ${\bf u}$ is the axis of rotation. Now it can be shown that conjugating $g$ by $h$ gives $$hgh^{-1}{\bf v}=\cos \alpha \,{\bf v}+(1-\cos \alpha)((h{\bf u})\cdot {\bf v})h{\bf u}+\sin \alpha \,(h{\bf u})\times {\bf v}.$$
This is a rotation by $\alpha$ about axis $h{\bf u}$.