Let $K$ be a field with an absolute value $v$ (not necessarily complete). Let $L$ be a finite Galois extension of $K$. Then, for any $\sigma \in Gal(L/K)$ and an absolute value $w$ on $L$ extending $v$, define the absolute value $\sigma w$ as $|a|_{\sigma w} = |\sigma^{-1}(a)|_w$. This gives a group action $Gal(L/K) \rightarrow ($ extensions of $v$ to $L)$.
Is this action transitive?
My attempt is : Let $L = K(a)$ for some $a \in L$ with minimal polynomial $f$. Then, extensions of $v$ to $L$ correspond to irreducible factors of $f$ in $\hat{K}$. Since $L/K$ is Galois, $f$ splits in $L$. The Galois group also acts transitively on the roots of $f$, so for any two roots $a_i$ and $a_j$ corresponding to irreducible factors $f_i$ and $f_j$, there is a $\sigma \in Gal(L/K)$ sending $a_i$ to $a_j$.
However, I'm not sure how to show that $\sigma$ also sends $|.|_i$ to $|.|_j$. I'm also stuck on relating $\sigma$, which acts on $L$, to $f_i$, which are polynomials in $\hat{K}[x]$. They are in different fields and I don't understand how to make a connection between them.
The answer to the question is yes, and it does not require using formulas for the extensions of the extended absolute values in terms of factoring a polynomial over the completion of the base field. The key idea to prove transitivity is to argue by contradiction: if some extended absolute value is not in the Galois orbit of another extended absolute value, we'll create an element in the base field whose absolute value is equal to $1$ by one method and is less than $1$ by another method.
The motivation for the proof below is the proof of transitivity of the Galois action on the prime ideals over a common prime number in a Galois extension of $\mathbf Q$. That proof relies on the Chinese remainder theorem for congruences modulo different prime ideals, and in place of that we'll use the analogue of the Chinese remainder theorem for absolute values: every finite set of inequivalent absolute values on a field is independent. What I did to get the proof below was start with the proof for Galois transitivity on prime ideals and translate it into the language of absolute values.
Theorem Let $E/F$ be a finite Galois extension and $|\,\cdot\,|$ be a nontrivial non-Archimedean absolute value on $F$. Assume there is an extension $|\cdot|_E$ of $|\cdot|$ to an absolute value on $E$.
(i) For $\sigma \in {\rm Gal}(E/F)$, $|x|_\sigma := |\sigma(x)|_E$ for $x \in E$ is an absolute value on $E$ extending $|\cdot|$, and if $|\cdot|_\sigma$ is equivalent to $|\cdot|_\tau$ where $\sigma, \tau \in {\rm Gal}(E/F)$, then $|\cdot|_\sigma = |\cdot|_\tau$.
(ii) Every absolute value on $E$ extending $|\cdot|$ is $|\cdot|_\sigma$ for some $\sigma \in {\rm Gal}(E/F)$
This theorem shows that if there are extensions of $|\cdot|$ to an absolute value of $E$, then all of them come from one of them by using pre-composition of one with elements of ${\rm Gal}(E/F)$. It can be proved that $|\cdot|$ extends to an absolute value on $E$ in at least one way.
Proof. (i): It is left to the reader to check that $|\cdot|_\sigma$ is an absolute value on $E$ extending $|\cdot|$. If $|\cdot|_\sigma$ and $|\cdot|_\tau$ are equivalent on $E$ then there is $t > 0$ such that $|x|_\sigma = |x|_\tau^t$ for all $x \in E$, so $|\sigma(x)|_E = |\tau(x)|_E^t$. When $x \in F$ this becomes $|x| = |x|^t$. Since $|\cdot|$ is nontrivial, there is an $x \in F$ such that $|x| \not= 0$ or $1$, so $t = 1$. Thus $|\sigma(x)|_E = |\tau(x)|_E$ for all $x \in E$.
(ii): We will argue by contradiction. Assume there's an absolute value $|\cdot|_E'$ on $E$ extending $|\cdot|$ such that $|\cdot|_E'$ is different from all $|\cdot|_\sigma$ for $\sigma \in {\rm Gal}(E/F)$. We're going to use this to construct $c \in E$ such that $|c| = 1$ and $|c|< 1$, which would be a contradiction.
Since $|\cdot|_E'$ restricts to $|\cdot|$ on $F$, just like all $|\cdot|_\sigma$, from $|\cdot|_E' \not= |\cdot|_\sigma$ for all $\sigma$ we get $|\cdot|_E'$ is inequivalent to all $|\cdot|_\sigma$ by the same argument as in (i). Inequivalent absolute values are independent, so using that with $|\cdot|_E'$ and representative absolute values among the inequivalent $|\cdot|_\sigma$, there is $x \in E$ such that $$ |x-1|_\sigma < 1 \ {\rm for \ all } \ \sigma \in {\rm Gal}(E/F) \ {\rm and } \ |x|_E' < 1. $$ (Note we quantify here over all $\sigma$, not just representatives for inequivalent $|\cdot|_\sigma$, because equivalent absolute values $|\cdot|_\sigma$ are in fact equal by (i).) Since $|x-1|_\sigma = |\sigma(x-1)|_E = |\sigma(x) - 1|_E$, rewrite the above displayed inequalities as $$ |\sigma(x)-1|_E < 1 \ {\rm for \ all } \ \sigma \in {\rm Gal}(E/F) \ {\rm and } \ |x|_E' < 1. $$ Because $|\cdot|$ is non-Archimedean, so is $|\cdot|_E$, so $|\sigma(x) - 1|_E < 1 \Rightarrow |\sigma(x)|_E = 1$. Therefore $$ |{\rm N}_{E/F}(x)| = |{\rm N}_{E/F}(x)|_E = \prod_{\sigma \in {\rm Gal}(E/F)} |\sigma(x)|_E = 1. $$ By a different argument, we'll also show $|{\rm N}_{E/F}(x)| < 1$.
Set $$ f(T) = \prod_{\sigma} (T - \sigma(x)) = T^n + c_{n-1}T^{n-1} + \cdots + c_1T + c_0, $$ so $f(T) \in F[T]$ (its coefficients are fixed by ${\rm Gal}(E/F)$). Each $c_i$ is, up to sign, a sum of products of the $\sigma(x)$, so from $|\sigma(x)|_E = 1$ for all $\sigma$ we get $|c_i| = |c_i|_E \leq 1$ (because $|\cdot|_E$ is non-Archimedean). For each root $r = \sigma(x)$ of $f(T)$ we have \begin{align*} r^n + c_{n-1}r^{n-1} + \cdots + c_1r + c_0 = 0 & \Longrightarrow r^n = -(c_{n-1}r^{n-1} + \cdots + c_1r + c_0) \\ & \Longrightarrow |r|_E'^n \leq \max_{0 \leq i \leq n-1} |c_i|_E'|r|_E'^i \leq \max_{0 \leq i \leq n-1}|r|_E'^i \end{align*} since $|c_i|_E' = |c_i| \leq 1$. If $|r|_E' > 1$ then $\max_{0 \leq i \leq n-1} |r|_E'^i = |r|_E'^{n-1} < |r|_E'^n$, which contradicts the above bound on $|r|_E'^n$, so $|r|_E' \leq 1$. Thus $|\sigma(x)|_E' \leq 1$ for all $\sigma$, so $$ |{\rm N}_{E/F}(x)| = |{\rm N}_{E/F}(x)|_E' = \prod_{\sigma \in {\rm Gal}(E/F)} |\sigma(x)|_E' \leq |x|_E' < 1, $$ which contradicts $|{\rm N}_{E/F}(x)| = 1$.