A point in 3-space, say $p=(x,y,z)$, can have a pure quaternion associated with it in the form $xi+yj+zk$ also called $p$. A quaternion $r=a+bi+cj+dk$ determines a linear mapping $\mathbb{R^3} \rightarrow \mathbb{R^3}$. The quaternion product $rpr^{-1}$ is also pure and we can also think of it as a point in $\mathbb{R^3}$, of the form $x^{\prime}i+y^{\prime}j+z^{\prime}k$. I want to show that when $r \ne 0$, ($rpr^{-1})^{-1}=r^{-1}pr.$ We are free to choose $r$ to have norm 1 to make this analysis simpler.
Does anyone have have any ideas or know how to show this?
It is not possible to show this. Both the left and right inverse of $rpr^{-1}$ is $rp^{-1}r^{-1}$. To show this just solve $q_l(rpr^{-1})=1$ and $(rpr^{-1})q_r=1$, and note that $q_l=q_r$.