Here is what I got so far. Let $\rho: Gal(L/K)\rightarrow GL(V)$ be an Artin representation. Let $\frak{p}$ be a prime of $K$, $\frak{P}$ a prime of $L$ lying above $\frak{p}$ and denote by $D_\frak{P}$ and $I_\frak{P}$ the decomposition and inertia groups respectively. Let $V^{I_\frak{P}}=\{v\in V | \rho(\sigma)(v)=v \ ,\ \forall \sigma\in I_\frak{P}\}$. Then is not hard to show that $\rho$ induces a representation $$\rho^{'}:D_{\frak{P}}/I_{\frak{P}}\rightarrow GL(V^{I_{\frak{P}}})$$ $$\tau I_{\frak{P}}\mapsto \rho(\tau)|_{V^{I_{\frak{P}}}}$$ Recall that the frobenius $Frob_\frak{P}$ is defined as any element in the pre-image of $Frob_{k({\frak{P}})/ k({\frak{p}})}\in Gal(k({\frak{P}})/ k({\frak{p}}))$ under the surjective homomorphism $$D_{\frak{P}}\rightarrow Gal(k({\frak{P}})/ k({\frak{p}}))$$ Hence we may have multiple choices for such frobenius (when $\frak{p}$ is ramified). But all such choices are equal modulo inertia, hence the existence of the representation $\rho^{'}$ allows us to define $$F_{\frak{P}}(t):=det(1-t\rho^{'}(Frob_{\frak{P}}I_{\frak{P}})^{-1})=det(1-t\rho(Frob_{\frak{P}})|_{V^{I_{\frak{P}}}})^{-1}$$which does not depend on the choice of such frobenius. Now it remains to show that $F_{\frak{P}}$ is invariant under the action of $Gal(L/K)$, i.e that $F_{\frak{P}}=F_{\sigma(\frak{P})},\ \forall\sigma\in Gal(L/K)$. I am positive that this boils down to some sort of ''char. poly. of similar matrices coincide"-type argument but I can't seem to be able to make it precise in this setting. I end up with the equality $$F_{\sigma(\frak{P})}(t)=det(1-t\rho(\sigma Frob_{\frak{P}} \sigma^{-1})|_{V^{\sigma I_{\frak{P}}\sigma^{-1}}})^{-1}$$and I want this to be equal to $det(1-t\rho(Frob_{\frak{P}})|_{V^{I_{\frak{P}}}})^{-1}$. I think this essentially reduces to the following linear algebra problem:
Given a diagram $$W\xrightarrow{\text{g}^{-1}} W^{'}\xrightarrow{\text{f}} W^{'}\xrightarrow{\text{g}} W$$ of invertible linear maps, then is it true that $$det(1-tf^{'})=det(1-tf)$$ where $f^{'}:=gfg^{-1}$ If this is indeed true, what is the easiest way to see it?
Also if this is true then we get what we want by letting $W=V^{\sigma I_{\frak{P}} \sigma^{-1}}, W^{'}=V^{I_{\frak{P}}}, f=\rho(Frob_{\frak{P}}), g=\rho(\sigma)$