Let $L/L'/K/\mathbb{Q}_p$ be a tower of finite field extensions such that $L'/K$ is totally ramified and $L/K$ is its Galois closure. We suppose that $L/L'$ is unramified. Let $M/K$ be a Galois subextension of $L/K$ such that $L/M$ is unramified. Is necessarily $L=M$ ?
The answer seems to be yes if the ramification of $L'/K$ is tame. We then have $L'=K(\sqrt[e]{\pi_K})$, so the residual degree $f(L/K)=[L:L']$ is equal to the multiplicative order of $|k_K|\bmod e$, where $|k_K|$ is the order of the residue field. If we write $M/M_0/K$ with $M_0/K$ maximal unramified, then $M_0$ must contain the $e$-th roots of unity, so $|k_{M_0}|\equiv 1\bmod e$. We conclude that $f(L/K)$ divides $f(M_0/K)$, so they must we equal. Then $M=L$ by minimality of $L$.
In general, if $Gal(L/L')$ is generated by a Frobenius lift $\phi$, then $Gal(L/M)$ is generated by an element $i\phi^k$ with some $i\in I(L/K)$ and some $k$ dividing $f(L/K)$. The group $Gal(L/M)$ is normal in $Gal(L/K)$, so one can prove that $\phi$ commutes with $i$. Then, $L'\cap M$ is the subfield of $L$ fixed by $\phi$ and $i$ and $$Gal(L/L'\cap M)=\langle\phi\rangle\times\langle i\rangle.$$ Furtherfore, one can prove that the quotient $$Gal(M/L'\cap M)=(\langle\phi\rangle\times\langle i\rangle)/\langle i\phi^k\rangle$$ is cyclic generated by the image of the Frobenius $\phi$.
My idea was to try to prove that $M/L'\cap M$ is unramified, thus leading to a conclusion that $i$ is trivial, and then that $k=1$ by minimality of $L$, but I am not even sure if that's possible.