How many automorphisms $\mathbb{C}$ has?

Question

How many automorphisms the complex numbers field has?

Answer 1

I assume you are ignoring the topology and just asking how many field automorphisms there are (over $\mathbb{Q}$, which must be fixed).

You might take a look at these notes. In brief, if $B$ is a transcendence base of $\mathbb{C}$ over the ground field $\mathbb{Q}$, and if $\phi: B \to B$ is any bijection, then there is an extension of $\phi$ to an automorphism of $\mathbb{C}$. Now the transcendence base here has cardinality $c = 2^{\aleph_0}$, and the cardinality of the set of bijections on $B$ will be $c^c = 2^c$, so there are at least $2^c$ automorphisms of $\mathbb{C}$ as a field. But since the number of functions $\mathbb{C} \to \mathbb{C}$ is also $c^c = 2^c$, we see that there are at most that many field automorphisms. So $2^c$ is the answer.

Edit: There is a very neat argument for why the cardinality of the set of bijections on a set $B$ of infinite cardinality $\beta$ is $\beta^\beta = 2^\beta$. Of course, $\beta^\beta$ is an upper bound since this counts the number of functions $B \to B$. To see $2^\beta$ is a lower bound, partition $B$ however you like into $\beta$ many 2-element sets (thus forming an equivalence relation). Then, for each set $A$ of equivalence classes, define a bijection on $B$ so that it swaps the two elements of a class if the class belongs to $A$, and fixes the elements of a class if it does not. The number of such bijections is the number of such $A$, which is $2^\beta$, so this is a lower bound for the cardinality of bijections on $B$.