Let $X$ be a module. $X$ is said to be *quasi-injective* if every homomorphism $h:A\to X$ from any submodule $A\subseteq X$ has an extension to an endomorphism $\tilde{h}:X\to X$. 

A module $X$ is quasi-injective if and only if $X$ is invariant under all endomorphisms of its injective envelope $I(X)$ [Theorem 1.1, [JohnsonWong1961][1]]. The backward direction is quite clear. For the forward direction, suppose $X$ is quasi-injective and pick an endomorphism $\phi:I(X)\to I(X)$. Let $Y=X\cap \phi^{-1}(X)$. The restriction $\phi|_Y:Y\to X$ has an extension $f:X\to X$ by hypothesis, and $f$ has an extension (also call it $f$) $I(X)\to I(X)$. If $g:=\phi -f=0$ on $X$ we're done. Otherwise, $X\cap g(X)\neq \{0\}$ since $X$ is an **essential** submodule of $I(X)$, so there exists nonzero $u,v\in X$  such that $g(u) = v$. Then, $$\phi(u) = f(u) + v\in X \Rightarrow u\in Y \Rightarrow v=g(u)=0$$ a clear contradiction. 

For a Banach space $X$, quasi-injectivity is analogous to extensibility. However, the obvious analogy of endomorphism invariance may imply $X=\{0\}$. For an example, let $X=\ell^2$ and $C(M)$ be its injective envelope. $M$ is the Gleason cover (a.k.a. projective cover) of the Cantor set $\{-1,1\}^{\mathbb{N}}$ [Theorem 3.3, [CohenLacey1969][2]]. If a proper subspace of $C(M)$ is invariant under the isometries of $C(M)$, then it is $\{0\}$.

>**Question:** What could be the correct analogy of [Theorem 1.1, [JohnsonWong1961][1]] for extensible Banach spaces ?


  [1]: https://doi.org/10.1112/jlms/s1-36.1.260
  [2]: https://doi.org/10.1016/0022-1236(69)90019-6