Let $F/K$ be a finite Galois extension where $(K,v)$ is a valued field (i.e. $v$ is a valuation on $K$). Let $w_1,w_2$ be extensions of $v$ to $F$. Then, we have associated valuation rings $O_{w_1}$ and $O_{w_2}$. I need to show that there exists $\sigma \in \text{Aut}{(F/K)}$ such that $O_{w_2}=\sigma(O_{w_2})$.
According to my lecture notes, I can see that if we can show that $O_v^*=\bigcap_{\sigma\in\text{Aut}{(F/K)}}\sigma(O_{w_1})$ where $O_v^*$ is the integral closure of $O_v$ in $F$, then we are done. However, I cannot see why that equality holds.
If $x\in F$ is integral over $O_v$, then it is so over $O_w$ for any extension $w$ of $v$. Since $O_w$ is integrally closed we get $x\in O_w$.
Conversely, let $x\in\bigcap_{\sigma\in\text{Aut}{(F/K)}}\sigma(O_{w})$, that is, $x\in O_{w\sigma}$ for all $\sigma\in\text{Aut}{(F/K)}$. (Note that $\sigma(O_w)=O_{w\sigma^{-1}}$.) Then $w(\sigma(x))\ge0$ for all $\sigma\in\text{Aut}{(F/K)}$. This shows that the coefficients of the minimal polynomial of $x$ over $K$ are in $O_w\cap K=O_v$.