Let $g:D^n\rightarrow D^n$ be a hyperbolic isometry, so it is a mobius transformation. Then, $g$ can be expressed as a composition $T\circ R$ where $T$ is a translation such that $T(0)=g(0)$ and $R$ is a Euclidean isometry fixing the origin.
My attempt: Since $g$ is mobius, $g$ is either on $O(n)$ or $g=f\circ R$ where $f$ is in $O(n)$ and $R$ is a rotation about a euclidean sphere. Here $g$ sends $0$ to $0$.
What is quick way to see this?
If $T$ is a hyperbolic translation from $0$ to $g(0)$ (which exists because hyp space is homogeneous) then consider $T^{-1}g$. Here, $T^{-1}g(0)=0$ and so $T^{-1}g$ is conjugate to a rotation and rotation is euclidean isometry.
But why this conjugation preserves euclidean isometry?