I have to proof that the only (field) automorphism of $\mathbb{Q}(\sqrt d)$ fixing $\mathbb{Q}$ are $id$ and the conjugation $\sigma$.
I know for every such automorphism $\tau$ we have that $$\tau(0)=0 \\ \tau(1)=1 \\ \tau(-a)=-\tau(a) \\ \tau(a^{-1})=\tau(a)^{-1}$$ and it is easy to see that satisfy all these properties. Any other automorphism I think about always violates any of these rules. But this does not mean that there could any very special and weird automorphism I just can not think about
An automorphism must fix $\mathbb Q$, since it fixes $1$, and thus all integers, and thus all inverses of integers, and thus all products of integers and inverses of integers, which covers all rational numbers.
Now consider the polynomial $f=X^2-d$. If $f(\alpha)=0$, then $f(\tau(\alpha))=\tau(f(\alpha))=\tau(0)=0$, so any automorphism has to send roots of $f$ to roots of $f$. That is, $\sqrt d$ is sent to itself (resulting in the identity automorphism), or to $-\sqrt d$ (resulting in the conjugation).