$\DeclareMathOperator{\AGL}{\operatorname{AGL}}\DeclareMathOperator{\PGL}{\operatorname{PGL}}$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.

Also generically, I believe there are no automorphisms for $n\geq 5$ by the following argument: We require all $4$ element subsets to have distinct cross ratios and since automorphisms have to preserve the cross ratio, this means that all $4$ element subsets are preserved. But this implies that the automorphism is trivial.

5more comments