Since the ramification indices of a prime are the same in a Galois extension, the following proposition is likely to be true. If it is true, how do we prove it?
Proposition Let $K$ be an algebraic number field. Let $L$ be a finite Galois extension of $K$. Let $G$ be the Galois group of $L/K$. Let $\mathfrak{D}_{L/K}$ be the different. Then $\sigma(\mathfrak{D}_{L/K}) = \mathfrak{D}_{L/K}$ for every $\sigma \in G$.