What is the automorphism group of $mathbb P^1$ minus $n$ points (let’s say over an algebraically closed field of characteristic $0$ if it matters). I want to consider the removed points without order. I can do small cases by hand but it seems hard in general and it seems to depend on which points are removed.

Here’s what I have thought about so far:

- $n = 0$: The automorphism group of $mathbb P^1$ is $PGL_2(k)$
- $n = 1$: The automorphism group of $mathbb A^1$ is AGL(1).
- $n = 2$: The automorphism group of $mathbb G_m$ is $mathbb Z/2 ltimes k^times$.
- $n = 3$: Since $PGL_2$ acts three transitively, it doesn’t matter which points we remove. Any automorphism of $mathbb P^1 – {0,1,infty}$ will extend to an automorphism of $mathbb P^1$ fixing ${0,1,infty}$ as a set and is determined by what it does to this set. We get all of $S_3$ in this case.
- $n = 4$: Any automorphism has to preserve the cross ratio and every permutation that does so is obtainable. So we get the Klein $4$ group – $mathbb Z/2 times mathbb Z/2$ in the generic case.

I don’t know what happens for $n geq 5$. We can get non trivial automorphisms for large $n$ by doing the following: Pick a finite subgroup of $PGL_2(k)$ (these are classified) and pick any finite subset of $mathbb P^1$ and remove the entire orbit of this set by the finite subgroup.