For $T(z)=1/(1-z),$ find $M\in \operatorname{Aut}(\Sigma)$ $(\Sigma=\mathbb{C}\cup \{\infty\},$ the Riemann sphere) such that $MTM^{-1}=ze^{2i\pi/3}.$ I know that $ze^{2i\pi/3}$ is a rotation of $\Sigma$ by an angle $2\pi/3$ about the vertical axis through $0$ and $\infty$ and also that a Moebius function is determined by three points. This two facts are helpful?
 A: $$
\begin{array}{cc|cc}
z \mapsto ze^{2\pi i/3} & & & z \mapsto \dfrac 1 {1-z} \\[10pt]
\hline
\begin{array}{ccccc} 1 & & \longrightarrow & & e^{2\pi i/3} \\ & \nwarrow & & \swarrow \\ & & e^{4\pi i/3} \end{array} & \quad & \quad & \begin{array}{ccccc} 0 & & \longrightarrow & & 1 \\ & \nwarrow & & \swarrow \\ & & \infty\end{array}
 \\[4pt] {} \end{array}
$$
Say you start with $1,\,\,e^{2\pi i/3},\,\, e^{4\pi i/3}$ respectively. Then you feed them into $M^{-1}$ getting $0,\,\,1,\,\,\infty$ respectively. Then you apply $T,$ transforming them to $1,\,\,\infty,\,\,0$ respectively. Then push them through $M,$ getting $e^{2\pi i/3},\,\, e^{4\pi i/3},\,\, 1$ respectively.
So you want $M^{-1}(1)=0,$ $M^{-1}(e^{2\pi i/3}) = 1,$ $M^{-1}(e^{4\pi i/3})=\infty.$
To get $M^{-1}(1)=0,$ you need $M^{-1}(z) = \dfrac{\cdots(z-1)}{\cdots}.$
Then to get $M^{-1}(\infty)= e^{4\pi i/3}$ you need $M^{-1}(z) = \dfrac{e^{4\pi i/3}(z-1)}{z - \cdots}.$
Then to get $M^{-1}(e^{2\pi i/3}) = 1,$ put $e^{2\pi i/3}$ in place of $z$ in that last expression and set it equal to $1$ and see what you need to put in place of $\text{“} \cdots \text{”}$ in order to get that result.
