Inertial degree of two primes on Galois extension is the same

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Let $K \subset L$ be a field extension of number fields, $\mathcal{O}_L$ be the ring of integers of $L$, and $\mathfrak{p}$ a prime ideal of $\mathcal{O}_K$. Then $\mathfrak p O_L$ is a prime ideal of $O_L$, and since rings of integers are Dedekind domains it has a decomposition into primes of $L$ containing $\mathfrak{p}$: $$ \mathfrak{p} \mathcal{O}_L = \mathfrak P_1^{e_1} \cdots \mathfrak P_g^{e_g},$$ where the integers $e_i$ are the ramification indexes of each $\mathfrak P_i$ and the degree $f_{\mathfrak P_{i }| \mathfrak{p}}$ of the extension $\mathcal{O}_K/\mathfrak{p} \subset \mathcal{O}_L/\mathfrak{\mathfrak P_i}$ is the inertial degree of $\mathfrak P_i$.

I'm trying to solve an exercise that asks to prove that for an automorphism $\sigma \in \text{Gal }(L/K)$ we have $e_{\mathfrak P_i} = e_{\sigma(\mathfrak P_i)}$ and $f_{\mathfrak P_{i }| \mathfrak{p}} = f_{\sigma(\mathfrak P_{i })| \mathfrak{p}}$. I could prove it for the ramification index, but i have no idea how to prove the inertial degree property.