TITLE: endomorphism of factor: can it be idempotent up to congugacy?
QUESTION [7 upvotes]: 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.

REPLY [7 votes]: 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.