This problem corresponds to the exercise 4.39 in Silverman's book "The Arithmetic of Dynamical systems":
Let $\alpha \in \mathbb{C}^*$, $d \geq 1$ and consider the rational function with degree $d$ $$\phi(z) = \left( \frac{z+1}{z-1} \right)^d.$$
Let $Aut(\phi) = \{ f \in PGL_2(\mathbb{C}): f \phi f^{-1} = \phi \}$ where the operation considere there is the composition of functions. Then $Aut(\phi) = \{ id \}$, i.e., the identity Mobius transformation.
MY ATTEMPTS: I have tried different things.
Of course the condition can also be seen as $f \phi = \phi f$. Putting a generic $f(z) = \frac{az+b}{cz+e}$ and making equations for $a,b,c,e$ does not really help me because the exponent $d$ that appears in $\phi$.
Evaluating at "nice" points. There are no nice points in my opinion, because when you try to make $f \phi$ nice, $\phi f$ is ugly, and nothing good arises.
Observing that $\phi$ is multiplying a Mobius transformation $\psi = \frac{z+1}{z-1}$ $d$ times. I could not get anything from this idea, but maybe it can inspire someone else.
Any help will be appreciated.