Here is a more elementary proof of the fact that there are no nontrivial measure preserving systems with the property that for all measurable $A,B$ there is some $N$ with $\mu(T^{-n}A\cap B) = \mu(A)\mu(B)$ for all $n \ge N$.
Take any $B$ with $0 < \mu(B) < 1$. Let $N_1$ be such that $\mu(B \cap T^{-N_1} B) = \mu(B)^2$. With $N_1,\dots,N_k$ chosen, choose $N_{k+1} > N_k$ such that $$\mu\left(\bigcap_{j \in \Sigma_k} T^{-N_j}B \cap T^{-N_{k+1}} B\right) = \mu(B)^{1+|\Sigma_k|}$$ for each $\Sigma_k \subseteq \{0,N_1,\dots,N_k\}$. Let $$A = \bigcup_{k \ge 1} [B \cap T^{-N_1} B \cap \dots \cap T^{-N_{k-1}}B \cap T^{-N_k}B^c \cap T^{-N_{k+1}}B].$$ Then, since the union is a disjoint one, $$\mu(A) = \sum_{k \ge 1} \mu(B)^{k+1}(1-\mu(B)) = \mu(B)^2,$$ so $$\mu(A)\mu(B) = \mu(B)^3,$$ while for $j \ge 3$, $$\mu(A \cap T^{-N_j} B) = \sum_{k=1}^{j-2} \mu(B)^{k+2}(1-\mu(B))+\mu(B)^j(1-\mu(B))+\sum_{k=j+1}^\infty \mu(B)^{k+1}(1-\mu(B))$$ $= \mu(B)^3+(1-\mu(B))^2\mu(B)^j.$