Two questions about automorphisms of the complex number field

Question

I have two questions regarding automorphisms of the complex number field. Suppose $f$ is an automorphism of the complex number field, such that $f(r)=r$ for every real number $r$. Must $f$ be either the identity or complex conjugation? That is to say, $f$ can't be a wild automorphism, right? My second, more general question is, if $f(\mathbb{R}) = \mathbb{R}$, that is, if the image of the set of real numbers is the set of real numbers, must $f$ be either the identity or complex conjugation?

Answer 1

Yes, every automorphism fixing each real number is either the identity or conjugation. This is because $i$ and $-i$ are the only possible automorphism images of $i$, and if $f$ is an automorphism then $f(a+bi)=f(a)+f(b)f(i)$. So we have two options satisfying $f(a)=a$ and $f(b)=b$, namely $f(a+bi)\in\{a+bi, a-bi\}$.More generally, the rule $$f(a+bi)=f(a)+f(b)f(i)$$ tells us that any $g:\mathbb{R}\rightarrow\mathbb{R}$ extends to at most two automorphisms of $\mathbb{C}$.

Towards your second question, suppose $f$ is an automorphism such that $f(a)<f(b)$ for some reals $a>b$ (note that this notation implies $f(a),f(b)\in\mathbb{R}$ for "$<$" to make sense). Consider the real number $c:=\sqrt{b-a}$. We have $c^2=b-a$ and so $f(c)^2=f(b)-f(a)$, but this means that $f(c)$ is not real. So your second question has a positive answer as well.