This is a question in the theory of random dynamical systems.
Let $(X,d)$ be a compact metric space, let $(I,\mathcal{I},\nu)$ be a probability space, and let $(f_\alpha)_{\alpha \in I}$ be an $I$-indexed family of continuous functions $f_\alpha:X \to X$ such that the map $(\alpha,x) \mapsto f_\alpha(x)$ is jointly measurable. Suppose we have a measurable function $a:I^\mathbb{N} \to X$ with the property that for $\nu^{\otimes \mathbb{N}}$-almost every sequence $(\alpha_1,\alpha_2,\alpha_3,\ldots) \in I^\mathbb{N}$,
$\ \ \ \ f_{\alpha_1}(a(\alpha_2,\alpha_3,\ldots)) \ = \ a(\alpha_1,\alpha_2,\alpha_3,\ldots)$.
Let $\rho$ be the image measure of $\nu^{\otimes \mathbb{N}}$ under $a$.
Is it necessarily the case that for $(\rho \otimes \rho \otimes \nu^{\otimes \mathbb{N}})$-almost every $(x,y,\alpha_1,\alpha_2,\ldots) \in X \times X \times I^\mathbb{N}$,
$\ \ d(f_{\alpha_n} \circ \ldots \circ f_{\alpha_1}(x),f_{\alpha_n} \circ \ldots \circ f_{\alpha_1}(y)) \to 0 \ \textrm{as} \ n \to \infty\,$?
[The papers which I have seen never seem to work on the basis that the answer is "yes" - but I don't know if this is because there are known counterexamples, or if this is because the answer is not known.]
Remark: It is known that for $\nu^{\otimes \mathbb{N}}$-almost all $\omega \! = \! (\alpha_1,\alpha_2,\alpha_3,\ldots) \in I^\mathbb{N}$, for every $\varepsilon>0$, if we let $B_{\omega,\varepsilon}$ denote the ball of radius $\varepsilon$ about $a(\omega)$ then
$\ \ \ \rho((f_{\alpha_1} \circ \ldots \circ f_{\alpha_n})^{-1}(B_{\omega,\varepsilon})) \to 1 \ \textrm{as} \ n \to \infty$.
As a result, it is known that for every $\varepsilon>0$,
$\ \ \rho \otimes \rho \otimes \nu^{\otimes n}( \, (x,y,\alpha_1,\ldots,\alpha_n) \, : \, d(f_{\alpha_n} \circ \ldots \circ f_{\alpha_1}(x),f_{\alpha_n} \circ \ldots \circ f_{\alpha_1}(y)) < \varepsilon \, ) \, \to \, 1$
as $n \to \infty$.