Theorem. The only field automorphism of $\mathbb{R}$ is the identity.
Proof. If $f \colon \mathbb{R} \to \mathbb{R}$ is a field automorphism, we want to prove that $f=\operatorname{id}$. The proof will be divided into three steps, following the treatment given in Makoto Kato’s answer in [1].
Step 1. $f \colon \mathbb{R} \to \mathbb{R}$ is order-preserving.
Let $x>0$ be a positive real number. Then there exists $y$ such that $x=y^2$, hence $f(x)=f(y)^2 > 0$. If $b > a$, then $b - a >0$ and so $f(b) - f(a)$ = $f(b - a) > 0$, i.e. $f(b) > f(a).$
Step 2. $f \colon \mathbb{R} \to \mathbb{R}$ fixes all elements of the subfield $\mathbb{Q}$.
Since $f(1)=1$ and every natural number n can be written as $n=1+1+ \ldots +1,$ we infer $f(n)=n$. This shows that $f$ preserves all elements of $\mathbb{N}$, and so it preserves all elements of the subring $\mathbb{Z}$. Consequently, it preserves all elements of the fraction field of $\mathbb{ℤ},$ which is $\mathbb{Q}$.
Step 3. We can now conclude by using the density of $\mathbb{Q}$ in $\mathbb{R}$. In fact, take any real number $x$ and let $q, \, r$ be two rational numbers such that $q < x < r$. By the previous two steps, we get $q < f(x) < r.$ Since $q$ and $r$ can be chosen to be arbitrarily close to $x$, it follows that $f(x)=x. \quad \square$
Remark. The situation is entirely different for the field $\mathbb{C}$ of complex numbers. Using an argument similar to the one above, it is possible to show that the only continuous automorphisms of $\mathbb{C}$ are the identity and the complex conjugation. However, admitting the Axiom of Choice it is possible to prove that there are infinitely many non-continuous automorphisms, see [2]. In fact, the continuous automorphisms of $\mathbb{C}$ are precisely those fixing the subfield $\mathbb{R}$.
[2] MSE412010-Wild automorphisms of the complex numbers
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