Let $F$ be an arbitrary field. Consider the field $F(x)$ consisting of all rational functions in the indeterminate $x$ with coefficients in the $F$.
Problem. Is there a reasonable way to describe $\operatorname{Aut}(F(x))$ of all field automorphisms of $F(x)$?
Note. We are not restricting attention to those field automorphisms $F(x)\to F(x)$ that fix $F$. That is an easier problem, since the group $\operatorname{Aut}(F(x)/F)$ consists of all automorphisms where $x$ is sent to $\frac{ax+b}{cx+d}$ where $a, b, c, d\in F$ satisfy $ab-cd\neq 0$.
Thoughts. There is probably a way to combine the automorphisms of $F$ itself and the automorphism above given by the fractional linear transformation. I wonder if there are other automorphisms of $F(x)$ that is genuinely different from such a construction. Of course, when $F=\mathbb{Q}$ and $F=\mathbb{F}_p$ are prime subfields, then any automorphisms of $F(x)$ must fix $F$ and we do not obtain anything new.
Suppose $F$ is the following subfield of the real numbers: $F=\mathbf{Q}(\pi)$. Then we can have many automorphisms of $F(x)$ sending $\pi$ to $(ax+b)/(cx+d)$.
When $F$ has elements that are both algebraic and transcendental over the prime subfield there will be more: we can take the automorphisms of the algebraic closure of prime subfield and do something like above.