Let $\phi : \mathbb Q(\sqrt2)\to \mathbb Q(\sqrt2)$ be an isomorphism. Show that if $\phi(1) = 1$, then $\phi(a) = a$ for all $a \in \mathbb Q$

Question

Let $\phi : \mathbb Q(\sqrt2) \to \mathbb Q(\sqrt2)$ be an isomorphism.

Show that if $\phi(1) = 1$, then $\phi(a) = a$ for all $a \in \mathbb Q$

My attempt:

$\phi(1) = \phi(a.a^{-1}) = \phi(a).\phi(a^{-1}) = 1$. Then $\phi(a) = \phi^{-1}(a^{-1})$.

I'm stuck here. Can anyone tell me how to continue this?