Let $X$ be a curve over a number field $K$ of genus $g\geq 2$. Does there exist an integer $d$ such that $X$ has infinitely many rational functions (i.e., finite morphisms $f:X\to \mathbf{P}^1_K$) of degree $d$?
If yes, can we choose/bound $d$ in terms of $K$ and $g$?
Note. This is not the same question as in the title. The answer to the question in the title is in fact negative as the example below shows for $d=2$.
Note: I want rational functions to be really different. So we mod out by the action of the automorphism group of $\mathbf{P}^1_K$. That is, $f$ and $\sigma\circ f$ are the same if $\sigma$ is an automorphism of $\mathbf{P}^1_K$.
Example 1. We can not have $d=2$. Hyperelliptic maps are unique.
Question 2. How does the answer to this question change if we replace $K$ by an algebraically closed field $k$?
Example 2. Let $X$ be a general curve of odd genus $g\geq 3$ over an algebraically closed field. Then, it has an infinite number of (really different) gonal morphisms. In fact, this family is one-dimensional.
Idea. I think the question can be reduced to a question on $\mathbf{P}^1_K$. In fact, it suffices to show that there are infinitely many (really different) rational functions on $\mathbf{P}^1_K$ of some degree, say $3$. In fact, once you know this, composing some morphism $f:X\to \mathbf{P}^1_K$ with such a rational function gives infinitely many rational functions of degree $d \leq 3 \deg f$. The only problem is then finding (in a controlled way) a morphism $f:X\to \mathbf{P}^1_K$. This is a hard problem, but let's allow finite base change if necessary...