@fedja: Could you please explain how you proved that cycles of any length are impossible if $A\neq I$? (Even though it's not a solution, it might give more insights on the problem...)

@NawafBou-Rabee: Yes, of course replacing the principal square root with the Cholesky square root makes the problem trivial. However the principal square root has remarkable properties that the Cholesky factor does not have (the most obvious one is that the principal square root is positive (semi)definite). In the problem I'm investigating, I need such properties.

@FedorPetrov: Yes, still it's not straightforward to me to find a pair of unit trace $X_0$, $X_1$ satisfying the above constraint for $A\neq I$. (In case of diagonal $X_0$, $X_1$ it's not possible, I would say).

@FedorPetrov: Since $A>0$ it looks improbable to me, but I could be wrong.

Yes, indeed, I'm wondering whether this still holds true with strict inequality.

Yes, I just edited the question in order to specify it.

It's a long story, but basically I would like to show that the iteration $(\star)$ has no invariant trajectories in the set of positive definite trace one matrices if $A\neq I$

Unfortunately, yes