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# endomorphism of factor: can it be indempotentidempotent up to congugacy?

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 non-invertiblenot 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.

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# endomorphism of factor: can it be indempotent up to congugacy?

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 non-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.