Fixed points of automorphism over quaternions

Question

I know that for any $y\in\mathbb{H}\setminus\{0\}$, the map $\rho_y:\mathbb{H}\to\mathbb{H}$, $\rho_y(x):=y\overline{x}y^{-1}$ is an automorphism. I would like to know, for fixed $y$, which are the fixed points of $\rho_y$. I was looking for something of the form $\{x\in\mathbb{H}: <x,y>=0\}$. Any help is appreciated.

Answer 1

Multiplying on the right with $y$, we see that $x\in \mathbb{H}$ is a fixed point of $\rho_y$ iff $y\overline{x}=xy$.

Case 1.

$\overline{y}=-y$, i.e. the real part of $y$ vanishes so $y$ is pure imaginary. Then \begin{align*} y\overline{x}&=xy\\ \iff -\overline{y}\,\overline{x}&=xy\\ \iff xy+\overline{xy}&=0\\ \iff x\cdot y&=0 \end{align*} as $x\cdot y=\frac{1}{2}(xy+\overline{xy})$. So, in this case, the fixed points are just quaternions orthogonal to $y$.

Case 2.

$\overline{y}\ne -y$, i.e. the real part of $y$ does not vanish, $y$ is not pure imaginary. In this case, suppose that $x$ is a fixed point, so $y\overline{x}=xy$. Then, conjugating this equation, $x\overline{y}=\overline{y}\,\overline{x}$ also. Then

\begin{align*} (y+\overline{y})(x-\overline{x})&=(y+\overline{y})x-(y+\overline{y})\overline{x}\\ &=x(y+\overline{y})-(y+\overline{y})\overline{x}\\ &=xy+x\overline{y}-y\overline{x}-\overline{y}\,\overline{x}\\ &=0 \end{align*} so $x=\overline{x}$. Hence, in this case, the fixed points are just the real numbers.