ag.algebraic geometry – What is the automorphism group of the projective line minus \$n\$ points?

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.