[Math] Showing $\text{Aut}(\mathbb{Q})$ is the trivial group and calculating $\text{Aut}(\mathbb{Q}(\sqrt{2}))$

Question

For a field $K$ let $\text{Aut}(K)$ denote the group of all automorphisms $f:K\to K$.

How can one show that $\text{Aut}(\mathbb{Q})$ is the trivial group and how to calculate $\text{Aut}(\mathbb{Q}(\sqrt{2}))$ ?

Answer 1

Hint: If $f$ is an automorphism of $\mathbb{Q}$, $f$ has to map every element to itself. (why?)

If $f$ is an automorphism of $\mathbb{Q}(\sqrt{2})$, $f$ fixes every element of $\mathbb{Q}$, meaning $f$ is determined by what it maps $\sqrt{2}$ to.