Iteration prime times of a rational map is Mobius function. Then the rational function form a group of order $1,p$ or $\infty$

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Consider a rational function $g=\frac{P(x)}{Q(x)}:S^2\to S^2$ and let $f$ be any Mobius function on $S^2$ where $S^2$ denotes Riemann sphere.

Let $p$ be a prime$\geq 2$. If $g^p=f$, then $g$ has order $1,p,\infty$ by considering the group formed by $(g)$ where multiplication is composition obviously.

Q1. Is this statement true in general?

I could show for concrete case like 5 being the case by explicit computation.

Q2. How do I show the general statement? I would love to have hints rather than solutions.