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
