Let $(F,v)$ be a valued field, with two extensions $(E_1,w_1)$ and $(E_2,w_2)$, i.e. $E_1$ (resp. $E_2$) extends $F$ as a field and $w_1$ (resp. $w_2$) extends $v$. We call $E_1$ and $E_2$ analytically equivalent if there exists an $F$-isomorphism of field $i\colon E_1\to E_2$, satisfying for all $a\in E_1$, one has $w_2(i(a))=w_1(a)$.
Now if $(K_1,v_1)$ and $(K_2,w_2)$ are two extensions of $(F,v)$, satisfying $(K_1,v_1)$ (resp. $(K_2,v_2)$) is analytically equivalent to an extension of $F$ contained in $(K_2,v_2)$ (resp. $(K_1,v_1)$). Can we conclude that $(K_1,v_1)$ is isomorphic to $(K_2,v_2)$ as valued field?
The analytic equivalence condition implies that $K_1$ and $K_2$ have same value group and residue field (contains each other), but it does not implies theirself are the same. On the other hand, I can't find a counterexample.