If unit quaternions $q_1,q_2$ satisfy $q_1^2=q_2^2=-1$, then the map $S^3\to S^3$, $x\mapsto q_1xq_2^{-1}$ has a fixed point Let $\Bbb H$ denote the quaternion algebra. We can identify $S^3$ as the subset of unit quaternions.
For fixed $q_1,q_2\in S^3$, consider the map $\phi:S^3\to S^3$ defined by $\phi(x)=q_1xq_2^{-1}$. How can we show that if $q_1^2=q_2^2=-1$ (so that $\phi^2=\text{id}$) then $\phi$ has a fixed point?
(This is claimed in a lecture note that I was reading, contained in the sketch of proof of the statement that if $G$ is a subgroup of $SO(4)$ of order $2$ acting freely on $S^3$, then $G=\{\pm \text{id}\}$.)
 A: The map $\phi:x\mapsto q_1x\overline q_2$ is ($\Bbb R$-)linear and as $\phi^2={\rm id}$, its minimal polynomial $m_\phi$ divides $(X^2-1)$.
If $m_\phi=(X-1)$ or $(X^2-1)$, then we're ready because $1$ is an eigenvalue of $\phi$.
The only remaining case is $m_\phi=(X+1)$, i.e. $\phi(x)=-x$.
But in that case $q_1\overline q_2=\phi(1)=-1$ implying $q_2=-q_1$, so
$$q_1x\overline q_1=-\phi(x)=x$$
for all $x\in\Bbb H$, or, $q_1x=xq_1$, that is, $q_1$ is in the center of $\Bbb H$, therefore $q_1\in\Bbb R$, namely $q_1=\pm1$, contradicting to $q_1^2=-1$.
A: In fact, we can characterize the set of fixed points geometrically.
The quaternion square roots of $-1$ are the unit vectors, I'll call them $\mathbf{u}$ and $\mathbf{v}$. The solutions to the equation $\mathbf{u}x\overline{\mathbf{v}}=x$ are the solutions to $x\mathbf{v}x^{-1}=\mathbf{u}$. If $x=\exp(\theta\mathbf{w})$ then this says $\mathbf{v}$ is obtained from $\mathbf{u}$ by rotating it around $\mathbf{w}$ by $2\theta$. Let's consider the case when $\mathbf{u},\mathbf{v}$ are parallel. If $\mathbf{v}=\mathbf{u}$ then $\mathbf{w}=\mathbf{u}$ too and $\theta$ is arbitrary, in other words $x\in\mathbb{R}[\mathbf{u}]=\mathrm{span}\{1,\mathbf{u}\}$. If $\mathbf{v}=-\mathbf{u}$ then $x$ anticommutes with $\mathbf{u}$, which means $x$ has no scalar part and must be a vector perpendicular to $\mathbf{u}$.
In general, if $S^3$ acts on $S^2$ by conjugation, the "transporter" from $\mathbf{v}$ to $\mathbf{u}$ (the set of solutions $x$) is a coset of $\mathbf{v}$'s stabilizer, so is a circle. To parametrize that circle in $S^3$ it suffices to find orthogonal elements of it.
If $\mathbf{u},\mathbf{v}$ are not parallel we can pick two very nice rotation axes: the one $\mathbf{w}_1$ which is between them, and $\mathbf{w}_2$ which is perpendicular to them. That is, $\mathbf{w}_1$ is $\mathbf{u}+\mathbf{v}$ and $\mathbf{w}_2$ is $\mathbf{u}\times\mathbf{v}$, but normalized. The former axis requires an angle of $\theta=\pi$ and the latter requires the angle $\theta=\angle\mathbf{uv}$. Then
$$ x_1=\frac{\mathbf{u}+\mathbf{v}}{\|\mathbf{u}+\mathbf{v}\|}, \quad x_2=\exp\left(\frac{\theta}{2}\frac{\mathbf{u}\times\mathbf{v}}{\|\mathbf{u}\times\mathbf{v}\|}\right) $$
can be used to parametrize all $x=(\cos\phi)x_1+(\sin\phi)x_2$. Note $x_2$ can be written purely in terms of $\mathbf{u},\mathbf{v}$ using Euler's formula, half-angle formulas, the dot product $\cos\theta=\mathbf{u}\cdot\mathbf{v}$, and Pythagorean identity.
