It's about the existence of a generalization of the first isomorphism theorem for groups, for subfactors :

Let $(N \subset M)$ and $(N' \subset M')$ be irreducible inclusions of hyperfinite $II_1$ factors.

Let $\phi: M \to M'$ be a $W^*$-morphism with $\phi(N) = N'$ and $\phi|_N : N \to N'$ an isomorphism.

Question: Is there an isomorphism of subfactors: $(\phi^{-1}(N') \subset M ) \simeq (N' \subset \phi(M))$ ?

(in other words, is there $\psi: M \to \phi(M)$ a $W^*$-isomorphism, with $\psi(\phi^{-1}(N')) = N' $ ?)

**Example**: let $N=N'=R$, $M=R \rtimes G $, $M'=R \rtimes G'$ with $G$ and $G'$ finite groups.

Let $f: G \to G'$ be a group-morphism.

First isomorphism theorem for groups : $G /ker(f) \simeq im(f)$.

Let $\phi: R \rtimes G \to R \rtimes G'$ be the canonical $W^*$-morphism coming from $f$.

Then, $\phi^{-1}(N') = R \rtimes ker(f)$ and $\phi(M)=R \rtimes im(f)$, so:

- $(\phi^{-1}(N') \subset M ) = (R \rtimes ker(f) \subset R \rtimes G) \simeq (R \subset R \rtimes G /ker(f))$
- $ (N' \subset \phi(M)) \hspace{0.5cm} = \hspace{0.5cm} (R \subset R \rtimes im(f)) \hspace{0.6cm} \simeq (R \subset R \rtimes G /ker(f)) $

**Edit** (after Dave Penneys's answer): The $W^*$-morphisms are too strong, we need to find weaker maps for this generalization be relevant and this example, correct.