Assume $L/K$ is a normal field extension and that $v$ is a valuation on $L$ (not necessarily discrete), with valuation ring $O_v$.
Let $\alpha\in O_v$, $\alpha\not\in K$, and let $\alpha'$ be a conjugate of $\alpha$ over $K$. Do we have $v(\alpha) = v(\alpha')$? are there general conditions for this to happen? (e.g. v discrete?, v ultrametric? L complete with respect to $v$? other?)
My try: Let us restrict ourselves to a Galois extension $L/K$ of Galois group $G$ (I think the general case reduces to this case anyway). Then $\alpha' = \sigma \alpha$ for some $\sigma \in G$.
We have $v(\alpha) = v(\sigma\alpha)$ iif $v((\sigma\alpha) / \alpha) = 0$, that is, $\sigma \alpha / \alpha$ is a unit of $O_v$.
We have $N_{L/K}(\sigma \alpha / \alpha) = N_{L/K}(\sigma \alpha)/N_{L/K}(\alpha) = 1$. So, it would suffice to prove that if $x\in O_v$ fulfills $N(x) = 1$, then $x$ is a unit of $O_v$.
This property holds (for all $\alpha$ in all algebraic extensions $L \vert K$) if $O_v$ is a Henselian ring. As per e.g. the source quotes in the comment, we also have the following converse: If $v$ extends uniquely this to all (normal, finite) algebraic extensions $L\vert K$, then $O_v$ must be Henselian. (Certainly it does not in general suffice to demand it for one specific extension. E.g. for $L=K$ the property is empty.)