It's known that every injective Banach space is of the form $C(M)$ where $M$ is a compact, Hausdorff, extremally disconnected topological space.
Let $X$, $Y$ be two injective Banach spaces such that,
- There exists an into linear isometry $i:X\to Y$, and
- There exists an into linear isometry $j:Y\to X$.
Q: Does there exist a surjective isometry $X\to Y$?
Dually, if $M_X$, $M_Y$ are compact Hausdorff extremally disconnected spaces such that there exist surjective continuous maps $j^{*}:M_X\to M_Y$, $i^{*}:M_Y\to M_X$, then
$\textbf{Q}^*\textbf{:}$ Are $M_X$ and $M_Y$ homeomorphic?
Edit(2024-07-05): I just realized that the answer is negative. A counterexample is the pair $\ell^{\infty}$ and $C(M_0)$, where $M_0$ is the Gleason cover of the Cantor set.
It is an old result of James (http://doi.org/10.1090/S0002-9939-1955-0076302-0) that $C(K)$ contains an isometric copy of $\ell^{\infty}$ if $K$ is extremally disconnected. So, there is an isometry $i:\ell^{\infty}\to C(M_0)$.
Conversely, there exists an isometry $\ell^2\to\ell^{\infty}$ defined by $x\to (\langle x,f_n\rangle )_{n\in\mathbb{N}}$ where $(f_n)$ is a norming subset of the dual unit ball. The injective envelope of $\ell^2$ is $C(M_0)$. Thus, there exists an isometry $j:C(M_0)\to\ell^{\infty}$.
Lastly, $C(M_0)$ and $\ell^{\infty}$ are (isomorphic as Banach spaces, but) not linearly isometric, since otherwise $M_0$ and $\beta\mathbb{N}$ would be homeomorphic, one of which contains no open point sets and the other contains a dense subset of open points.