Define $\operatorname{Im} H := \{r+si+tj+uk \in H : r = 0\},\, \phi(si+tj+uk):=(s,t,u)$, and \begin{align} \operatorname{Im} H &\to \operatorname{Im} H \\ p &\mapsto qpq^{-1} \end{align}
Let $q \in H$ so that $|q|=1$.
Show that $\phi \circ T_q \circ \phi^{-1}:\mathbb{R}^3 \to \mathbb{R}^3$ belongs to $SO(3)$, the space of $3\times 3$-rotation matrices.
Then $$ \phi \circ T_q \circ \phi^{-1}(s,t,u)=\phi(T_q(\phi^{-1}(s,t,u)))=\phi(T_q(si+tj+uk))=\phi(q(si+tj+uk)q^{-1}. $$
This is where I'm stuck as $\phi$ takes quaternions and the argument is a matrix.
I guess I need to show then that $q(si+tj+uk)q^{-1}$ is an orthogonal matrix with determinant $1$?
Thanks for any answers!
Let $N\colon H\longrightarrow\Bbb R$ be the quaternionic norm. That is, $N(a+bi+cj+dk)=\sqrt{a^2+b^2+c^2+d^2}$. Then, if $(s,t,u)\in\Bbb R^3$, $\|\phi(si+tj+uk)\|=N(si+tj+u)$. I other words, $\phi$ preserves the norm. ANd therefore so does $\phi^{-1}$.
Note that $T_q$ is also norm-preserving: if $p\in\operatorname{Im}H$, then$$N(T_q(p))=N(qpq^{-1})=N(q)N(p)N(q)^{-1}=N(p).$$
But then, if $(s,t,u)\in\Bbb R^3$,$$\begin{align}\left\|\phi\bigl(T_q\bigl(\phi^{-1}(s,t,u)\bigr)\bigr)\right\|&=N\bigl(T_q\bigl(\phi^{-1}(s,t,u)\bigr)\bigr)\\&=N\bigl(\phi^{-1}(s,t,u)\bigr)\\&=\|(s,t,u)\|.\end{align}$$
So, $\phi\circ T_q\circ\phi^{-1}\in O(3)$. But the map$$\begin{array}{ccc}\operatorname{Im}H&\longrightarrow&O(3)\\q&\mapsto&\phi\circ T_q\circ\phi^{-1}\end{array}$$is continuous and its domain is connected. Therefore, its range is connected. Since the identity matrix belongs to the range, the range is a subset of $SO(3)$, which is the connected component of $O(3)$ to which the identity matrix belongs.