Let $X$ be a complex algebraic curve, assumed to be connected, smooth and complete. Let $f: X \rightarrow X$ be a surjective morphism. Define a *backward complete set for $f$* as a subset $S$ of $X$ such that $f^{-1}(S) \subset S$ (I am not sure if it is the standard terminology).



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> If $f$ has infinitely many finite backward complete sets, is $f$ necessarily an automorphism?
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A question on dynamics on complex algebraic curves
For a map $f:\mathbb P^1\to\mathbb P^1$ of degree $d\ge2$, there are three cases: (1) There are no finite backward invariant sets. (2) There is one such set consisting of a single point.Moving that point to infinity, we have $f(x)\in\mathbb{C}[x]$ is a polynomial of degree $d$. (3) There are two such points. Moving them to $0$ and $\infty$, the map $f$ has the form $f(x)=cx^{\pm d}$. 


This is a standard first result in complex dynamics. The proof is quite easy using the Riemann-Hurwitz genus formula, which for $f:\mathbb P^1\to\mathbb P^1$ says that
$$ 2d-2 = \sum\_{P\in\mathbb P^1} (e\_P(f)-1). $$
If $\{P\_1,\ldots,P\_n\}$ is a backward invariant set, then each $f^{-1}(P\_i)$ must consist of a single point, so the points in the set are totally ramified, i.e., $e\_{P\_i}(f)=d$. Hence
$$ 2d-2 = \sum\_{P\in\mathbb P^1} (e\_P(f)-1)
\ge \sum\_{i=1}^n (e\_{P\_i}(f)-1) = n(d-1).$$
This proves that $n\le2$, and a more careful analysis of the $n=1$ and $n=2$ cases yields the results mentioned earlier.


