Archimedean and Euclidean implies trivial automorphism group

Statement

Suppose $K$ is a Euclidean Archimedean field: a field that is both Euclidean and Archimedean. In other words, there exists a total ordering $\leq$ on $K$ making $K$ into an ordered field, such that $0\leq a$ if and only if $a$ is a square. Then, the automorphism group of $K$ is trivial.

Proof

Proof outline

1. Any field automorphism of $K$ must send squares to squares.
2. Any field automorphism of $K$ is also an automorphism of $K$ as an ordered field, i.e., it preserves the total ordering.
3. We now use the fact that the field automorphism fixes every rational number, combined with the fact that the rational numbers are dense in $K$.