Conjugating the Mobius Transformation $\mathcal M(z)=1/(z-2i)$

Question

Consider the Mobius transformation $\mathcal M:\textbf C^*\to\textbf C^*$ (where $\textbf C^*:=\textbf C\cup\{\infty\}$ denotes the extended complex plane/Riemann sphere) defined by: $$\mathcal M(z):=\frac{1}{z-2i}$$ I'm pretty sure this particular Mobius transformation $\mathcal M$ has only one fixed point in $\textbf C^*$, namely at $\mathcal M(i)=i$. And I'm also pretty sure that the collection of all Mobius transformations with exactly one fixed point is equal to the conjugacy class of the simple rightward unit Mobius translation $\mathcal T(z):=z+1$ (which has one fixed point at $z=\infty$) in the Mobius group. Therefore, I thought it should be possible to conjugate $\mathcal M$ by some intermediary Mobius transformation $\mathcal I(z)=\frac{az+b}{cz+d}$ such that: $$\mathcal T=\mathcal I\circ\mathcal M\circ\mathcal I^{-1}$$ However, I seem to be unable to find an intermediary Mobius transformation $\mathcal I(z)$ that successfully accomplishes this conjugation. Any ideas why?