Let $p$ be a prime and let $K$ be a finite extension of $\mathbb{Q}_p$. Suppose $L/K$ is a tamely ramified Galois extension. I want to show that if $\sigma$ is a lift of the Frobenius element of the Galois group of the residue field extension and $\tau$ is a generator of $\textrm{Gal}(L/E)$ where $E$ is the maximal unramified subextension of $L$, then $\sigma\tau\sigma^{-1} = \tau^q$ where $q$ is the order of the residue field of $K$.
I know that $L = E(\lambda^{1/e}$) where $\lambda$ is some uniformizer for $E$ and $e$ is the ramification index of $L/K$. If $\sigma(\lambda^{1/e})$ is another $e$-th root of $\lambda$, then I can show the desired relation holds. However, it seems to me that we can't be sure this is the case, because $\lambda$ may not be in $K$.