I want to show that the following map is in $SO(\mathbb{H})$. Let $A,B\in SU(2)$ define $F:H\rightarrow H$ by $F(h)=AhB^{-1}$.
Intuitively this makes sense since $SU(2)$ can be identified with the unit quaternions, so composing with a rotation on both sides should give a map that has determinant 1.
As far as I understand to show that $F$ is in $SO(\mathbb{H})$ we have to find a matrix $X$ in $SO(\mathbb{H})$ such that $F(h)=Xh$. But this does not seem possible.
If $A,B\in SU(2)$ then we have $A= \left(\begin{matrix}a&-\overline{b}\\b&\overline{a}\end{matrix}\right)$ and $B=\left(\begin{matrix}c&-\overline{d}\\d&\overline{c}\end{matrix}\right)$. For $h\in H$ (abusing notation) we have $h=\left(\begin{matrix}g&h\\-\overline{h}&\overline{g}\end{matrix}\right)$.
But $AhB$ cannot be equal to $Xh$ since, the first entry in $AhB^{-1}$ has contributions from $g,h,\overline{g},\overline{h}$ but the first entry in $Xh$ only has contributions from $g,\overline{h}$.
Question: Is trying to show that $F(h)=Xh$ the correct approach? If so where am I going wrong?
No, you don't have to find a matrix. If $A,B\in SU(2)$ and $h\in\mathbb H$, then$$\lVert AhB^{-1}\rVert=\lVert A\rVert.\lVert h\rVert.\lVert B^{-1}\rVert=\lVert h\rVert.$$So, $h\mapsto AhB^{-1}$ belongs to $O(\mathbb H)$. All that remains to be proved is that its determinant is $1$. But note that $SU(2)$ is connected and therefore, since your map is a continuous map from $SU(2)\times SU(2)$ into $O(\mathbb H)$, its image is connected. And its image contains $\operatorname{Id}$ (just take $A=B=\operatorname{Id}$). So, its image is a subset of the connected component of $\operatorname{Id}$ in $O(\mathbb H)$, which is $SO(\mathbb H)$.