Let $\mathbb{E}\supseteq\mathbb{F}$ be a field extension with $\dim_\mathbb{F}(\mathbb{E})<\infty$ and let $G\subseteq\mathrm{Aut}(\mathbb{E})$ be the subgroup of field automorphisms $\sigma:\mathbb{E}\to\mathbb{E}$ that fix $\mathbb{F}$. What is the easiest way to prove that $G$ is finite?
Edit: How about this? Every element of $\mathbb{E}$ is algebraic over $\mathbb{F}$. By induction we can write $$\mathbb{E}=\mathbb{F}[\alpha_1,\ldots,\alpha_n].$$ Let $\Omega_i$ be the (finite) set of roots of the minimal polynomial of $\alpha_i$ over $\mathbb{F}$. Then $G$ acts on the Cartesian product $\Omega_1\times \cdots \times\Omega_n$. Let $\mathcal{O}$ be the (finite) $G$-orbit of the element $$(\alpha_1,\ldots,\alpha_n)\in\Omega_1\times\cdots\times\Omega_n.$$ Note that the stabilizer is trivial since if $\sigma\in G$ fixes each $\alpha_i$ then it fixes any polynomial expression $f(\alpha_1,\ldots,\alpha_n)$, hence it fixes every element of $\mathbb{E}$. It follows from the orbit-stabilizer theorem that $\#G=\#\mathcal{O}$ is finite.