Orders of field automorphisms of $\mathbb{C}$

Question

It is well known that, given the axiom of choice, $\mathbb{C}$ has $2^{2^{\aleph_0}}$ field automorphisms. Except that the non-obvious automorphisms cannot be constructed explicitly, can we say more about them? Specifically, is there one $\sigma\in \mathrm{Aut}(\mathbb{C})$ with finite order $>2$ (that is, $\sigma^n = \mathrm{id}$ for some $n>2$ but $\sigma^2 \neq \mathrm{id}$) ? Does every finite order occur? Is there one $\sigma$ with infinite order?

Answer 1

Where are you stuck at constructing some $\sigma$ with infinite order?

If finite then the order is $2$, this is explained in https://kconrad.math.uconn.edu/blurbs/galoistheory/artinschreier.pdf .

With $\rho$ the complex conjugaison, every $\sigma \rho \sigma^{-1}$ has order $2$, this gives infinitely many examples.

There are also order $2$ automorphisms that are not of this form (for example the one of $\Bbb{C}((x))/\Bbb{R}((x))$ conjugated with an isomorphism $\Bbb{C}((x))\to \Bbb{C}$)