finding all automorphism of $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}(\sqrt{2})$ If we want to find all the automorphisms of $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}(\sqrt{2})$,
The generator is $\sqrt[4]{2}$,
We look at the minimal polynomial, in this case its $x^2 - \sqrt{2}$ (I'm not sure why we look at this)
Basically we know that an automorphism $\phi$ must map identity to identity,
$$\phi(a+b\sqrt[4]{2})\to a\pm b\sqrt[4]{2}$$
are the only automorphisms. I'm not quite sure why.
 A: Note that $\sigma$ is determined by where does it send $\sqrt[4]{2}$ to ?
As you have the minimal polynomial $x^2-\sqrt{2}$ with real roots $\pm \sqrt[4]{2}$, and you also know that every automorphism sends the roots of the minimal polynomial to its another root i.e., every automorphism permutes the roots of the minimal polynomial,  you have the following options:
\begin{align}
&\sqrt[4]{2}\to \sqrt[4]{2},~-\sqrt[4]{2}\to-\sqrt[4]{2}~~~(identity) \\
&\sqrt[4]{2}\to-\sqrt[4]{2},~-\sqrt[4]{2}\to\sqrt[4]{2}
\end{align}
Thus you can have at most  two automorphisms.
A: Whenever you have a degree $2$ algebraic element $b$ over a field $F$ of characteristic $0$, the minimal polynomial splits over $F(b)$.
Indeed, if the minimal polynomial is $x^2+px+q$, one root is $b$ and the other root is $-p-b=q/b$.
Therefore $F(b)$ is a Galois extension and the Galois group has order $2$. The nonidentity automorphism just swaps the roots.
In your case the minimal polynomial of $\sqrt[4]{2}$ over $\mathbb{Q}(\sqrt{2}\,)$ is $x^2-\sqrt{2}$. The nonidentity automorphism is $a+b\sqrt[4]{2}\mapsto a-b\sqrt[4]{2}$.
