For $A,X,Q \in \mathbb{R}^{n \times n}$, define $h(X) = A X A^T + Q$ and $ h^j(X)=\underbrace{{h(h(}...h}_{j\text{ times}}(X)))$.
If $X,Q$ are positive definite, $A\neq 0$ and for a certain integer $n \geqslant 2$, there is $h^n(X)>X$. Is it true that $h(X)>X$?
Here "$A>B$" means "$A-B$" is positive definite.
I've made a mistake. The examples provided by @Omnomnomnom is not a contradiction.
In my original proplem, $X$ is accually the unique positive definite solution of function $X = g(h^n(X))$. Where, the function $g(Y)$ has the following property: $Y>g(Y)$.
I don't know how this helps, but thought of writing it up since some one here might be able to extend it. We can write $$h^n(X)=A^nX(A^n)^T+\sum_{i=0}^{n-1}A^iQ(A^i)^T$$ where ($A^{0}=I$).