As mentioned on Wikipedia, let $L/K$ be a finite extension of number fields, and $\mathcal O_L$ and $\mathcal O_K$ the respective ring of algebraic integers. If we assume that $L/K$ is Galois, then we can consider the group action of the Galois group $G := \mathrm{Gal}(L/K)$ on the ring extension $\mathcal O_L/\mathcal O_K$ to show that $G$ acts transitively on the prime factors $\mathfrak P_1, \dots, \mathfrak P_r \subset \mathcal O_L$ of $\mathfrak p \mathcal O_L$, where $(0) \ne \mathfrak p \subset \mathcal O_K$ is some prime.
My question is: Why do we need $G$ to be the Galois group? The standard proof by way of contradiction, using the Chinese remainder theorem, (see e.g. here) would work for any group action with finite $G$ where $\mathcal O_L^G = \mathcal O_K$. One could even generalise this to any group action on finite Dedekind ring extensions $A \subseteq B$ and need not require any fields in the statement or proof at all.
Best Answer
The premise that $G$ is a finite group of $K$-automorphisms of $L$ such that $\mathcal O_L^G = \mathcal O_K$ implies $L^G = K$ and Artin proved that implies $L/K$ is Galois with Galois group $G$.
That is, if $G$ is finite group of field automorphisms of $L$ and we set $K = L^G$, then Artin showed $L/K$ is a finite Galois extension and the natural mapping $G \to {\rm Gal}(L/K)$ is an isomorphism. Artin's definition $K = L^G$ is, in your case, provable from your assumption that $\mathcal O_L^G = \mathcal O_K$.