Does any isomorphism of splitting fields automatically fix the base field?

Question

I was wondering whether it's true that if $L_1,L_2$ are splitting fields of $f\in \Bbbk[x]$ over $\Bbbk$ then any isomorphism between them as fields must fix $\Bbbk$. If this is true, why? Is there any intuition to be had?

Answer 1

No. For example, let $L_1=\mathbb Q(\sqrt[4]2,i)$ be the splitting field of $f(X) = X^4-2$ over $K=\mathbb Q(\sqrt[4]2)$. Let $\sigma$ be the automorphism of $L_1$ which sends $\sqrt[4]2\mapsto i\sqrt[4]2$ and fixes everything else.

Then $\sigma(L_1)$ is isomorphic to $L_1$ and is another splitting field of $f$ over $K$. But $\sigma$ does not even reduce to an automorphism of $K$, whilst $\sigma^2$ reduces to a non-trivial automorphism.