Centralizer of group of unit quaternions Consider the multiplicative group $\Bbb S^0(\Bbb H)$ of unit quaternions as a subgroup of $\mathrm O(\Bbb R^4)$. 
I wonder what would be the centralizer of that group, that is, the set of orthogonal matrices that commute with all elements of $\Bbb S^0(\Bbb H)$:
$$C(\Bbb S^0(\Bbb H)) = \{X\in\mathrm O(\Bbb R^4 \mid XT = TX\text{ for all $T\in\Bbb S^0(\Bbb H)$}\}.$$
I somehow believe that (a connected component of) the centralizer is isomorphic to $\Bbb S^0(\Bbb H)$, but I have no idea how to find this explicitly.
 A: Let $\phi:\mathbb{H}\rightarrow \mathbb{H}$ be an $\mathbb{R}$-linear map such that for every unit quaternion $q$ and every element $x\in \mathbb{H}$ we have (the centralizer condition)
$$\phi(q\cdot x) = q\cdot \phi(x)$$
You are looking for such $\phi$ with additional property that $\lvert\phi\left(x\right)\rvert = |x|$ for every $x\in \mathbb{H}$ (the isometry condition). 
Pick any nonzero quaternion $x$. Then $q = \frac{x}{|x|}$ is a unit quaternion and we have
$$\phi(x) = \phi\left(q\cdot |x|\right) = q\cdot \phi(|x|) = q\cdot |x|\cdot \phi(1) = x\cdot \phi(1)$$
Thus $\phi$ is given by the multiplication from the right by $\phi(1)\in \mathbb{H}$ and since it preserves the length of vectors, we derive that $|\phi(1)| = 1$. Therefore, $\phi$ is the multiplication from the right by unit quaternions.
So in your notation 
$$C\left(S^0(\mathbb{H})\right) \cong \left(S^0(\mathbb{H})\right)^{op}$$
is the opposite group to $S^0(\mathbb{H})$. Since for every group $G$ we have $G\cong G^{op}$ by virtue of a map $g\mapsto g^{-1}$, we deduce that your group is isomorphic with $S^0(\mathbb{H})$.
