Extension theorem for field automorphisms

Let's consider the field extension $$\mathbb Q\subset \mathbb Q(\sqrt 2)\subset\mathbb Q(\sqrt[4]{2})$$

and construct with it the $\mathbb Q$-Automorphisms of $\mathbb Q(\sqrt[4]{2})$. There is only one automorphism on $\mathbb Q$, namely the identity. So we get all $\mathbb Q$-Automorphisms of $\mathbb Q(\sqrt 2)$ by extending the identity on $\mathbb Q$. Since the two roots of the minimal polynomial of the primitive element are $\pm\sqrt 2$, there are the two extensions $$\phi_1:\sqrt 2\mapsto \sqrt 2\ \text{ and }\phi_2:\sqrt 2\mapsto -\sqrt 2$$ Now, if I try to reproduce this in order to obtain the $\mathbb Q$-Automorphisms of $\mathbb Q(\sqrt[4]{2})$ I encounter en error. The field extension is primitive with $\sqrt[4]{2}$ being a primitive element (maybe that's where I am wrong) so the extension theorem says that both $\phi_1$ and $\phi_2$ can be extended further onto $\mathbb Q(\sqrt[4]{2})$. But how can $\phi_2$ be extended if we always have $$\phi_2(\sqrt[4]{2})^2=\phi_2(\sqrt[4]{2})\phi_2(\sqrt[4]{2})=\phi_2(\sqrt 2)=-\sqrt 2?$$Clearly, $\mathbb Q(\sqrt[4]{2})$ is a subset of $\mathbb R$ so there is no proper image that would satisfy the equation above. I know that $\mathbb Q(\sqrt[4]{2})$ isn't a normal extension of $\mathbb Q$, but it is a normal extension of $\mathbb Q(\sqrt 2)$ so where is my misunderstanding of that theorem?

I'm assuming that you are reffering to the following theorem: "Let $\phi:F_1 \to F_2$ be a field isomorphism. Let $f_1 \in F_1[x]$ and $f_2 \in F_2[x]$ be obtained by applying $\phi$ to the coefficients of $f_1$. Then if $L_1, L_2$ are splitting fields of $f_1,f_2$ over $F_1,F_2$ respectively we have that there exists a field isomorphism $\overline\phi:L_1 \to L_2$ which restricts to $\phi$ on $F_1$"
In your case indeed $\mathbb{Q}(\sqrt[4]{2})$ is normal extension over $\mathbb{Q}(\sqrt{2})$ and it's also a splitting field of $f_1 = x^2 - \sqrt{2} \in \mathbb{Q}(\sqrt{2})[x]$. The problem with you applying the theorem is that the isomorphism you get will be from $\mathbb{Q}(\sqrt[4]{2})$ to a splitting field of $f_2$ over $\mathbb{Q}(\sqrt{2})$, not necessarily an automorphism of $\mathbb{Q}(\sqrt[4]{2})$. Indeed the later case isn't true in our case, as $f_2 = x^2 + 2 \in \mathbb{Q}(\sqrt{2})[x]$ and a splitting field of it is $\mathbb{Q}(i\sqrt[4]{2})$ and not $\mathbb{Q}(\sqrt[4]{2})$.