Say we have the following tower of fields: $\mathbb{Q} \subset \mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\sqrt[4]{5})$. For ease, we can write $\mathbb{Q} = F, \mathbb{Q}(\sqrt{5}) = E, \mathbb{Q}(\sqrt[4]{5}) = K$.
I know that $Aut_F(E) = \{id, \tau: \sqrt{5} \mapsto -\sqrt{5} \}$.
I am told that $\tau$ as defined is not an F-automorphism of $K$. Can someone help me figure out why? I am wondering what $\tau$ would have to do to $\sqrt[4]{5}$ in order to be an $F$-automorphism of $K$, and why it doesn't do what it's supposed to do. I don't think the following computation is correct (although it gives the right result), but I'm not sure how to fix it: $$\tau(\sqrt[4]{5}) = \tau(\sqrt{5}^{1/2}) = \tau(\sqrt{5})^{1/2} = -\sqrt{5}^{1/2}$$ where the second-to-last equality follows because $\tau$ is an automorphism. This does not describe a $K$-automorphism of $F$ because $-\sqrt{5}^{1/2} \not\in K$.
I know that $E$ is a Galois extension of $F$, and $K$ is not since $i \not\in K$. Could this be relevant, ie. is there a stronger result hiding here?