Let $M$ be a factor, and let $\phi:M\to M$ be an irreducible endomorphism
("irreducible" means that the relative commutant of $\phi(M)$ in $M$ is trivial).
Let's also assume that $\phi$ is not invertible.


Is it possible to have $\phi\circ \phi$ conjugate to $\phi$?  

In other words, is it possible to have an endomorphism $\phi$, and a unitary $u\in M$, such that $$\phi(\phi(x))=u\phi(x)u^\*,\quad\forall x\in M.$$


If this is possible, I would like to see an example.


---


*Note:* an answer to the above question would also settle [this question](https://mathoverflow.net/questions/41327/subfactor-of-finite-rank-but-infinite-index-is-this-possible).



endomorphism of factor: can it be idempotent up to congugacy?
This is not possible. If it were, then using the notation above, given any $x \in \phi(M)$, we would have $x u^\* = u^\* \phi(x)$, and $\phi(x) u = u x$. Hence, for any $x \in \phi(M)$ we have
$$
x u^\* \phi(u^\*) u^2 = u^\* \phi(x u^\*) u^2
$$
$$
= u^\* \phi(u^\*) \phi \circ \phi (x) u^2 = u^\* \phi(u^\*) u^2 x.
$$ `


Hence $u^\* \phi(u^\*)u^2 \in \phi(M)' \cap M = \mathbb C$ and so $\phi(u^\*) \in \mathbb C \cdot u^\*$. Then, for any $y \in M$ we would have that 
$$
\phi \circ \phi (y) = u \phi(y) u^\*
$$
$$
= \phi(u y u^\*).
$$
Since $\phi$ is injective we then have $\phi(y) = u y u^\*$, and hence $\phi$ is invertible.


If you don't require that $\phi(M)$ be irreducible then this is possible.


