Let $S =[0, 1]^2$ denote the unit square in $\mathbb R^{2}$. For any subset $A$ of $S$ let $A^{c}$ denote its complement in $S$, and $\overline{A}$ its closure in $S$. Given a measurable map $g: W \to V$ between measure spaces $(W, \mu)$, $(V, \phi)$, we say it is equally scaling if there exists some $r$ in $\mathbb R$ such that for any measurable set $K$ in $V$, we have $\phi(K) = r\mu(g^{-1}(K))$.
Question: Does there exist
a sequence of measurable sets $A_n$ in $S$ such that for every $n$ in $\mathbb N$, the closures $\overline{A_n}$, $\overline{A_n ^{c}}$ are homeomorphic to $S$ and have Lebesgue measure $\frac{1}{2}$,
and two sequences of equally scaling homeomorphisms $h_{n}:$ $\overline{A_n} \to $ $\overline{A_n ^c}$, $s_n:A^c \to S$ such that for any continuous function $f \in C^0 (S)$, $$ \lim_{n \to \infty} F_{n}...F_{0}(f) \to \int_S f$$ uniformly?
Where $F_{n}: C^0 (S) \to C^0(S) $ is defined by $$F_{n}(f) (x) = \frac{[fs_n^{-1}(x) + fh_n^{-1}s_n^{-1}(x)]}{2} $$
