Suppose $G$ is a finite group and $\{X_{i,j}\}_{i, j \in \mathbb{N}}$ are a set of i.i.d. random elements of $G$. Now suppose that $\{A_i\}_{i = 0}^\infty$ is a sequence of random elements of $G$ defined by the following relations:
$$P(A_0 = e) = 1$$ $$A_{n+1} = \Pi_{i = 1}^{ord{A_{n}}} X_{(n+1), i}$$
Here $\Pi$ stands for iterated group product, and $ord$ for the order of an element.
My question is:
Is it always true, that $\exists H \leq G$, such that $\forall g \in G$ $\lim_{n \to \infty} P(A_{n} = g) = \frac{I_H(g)}{|G|}$?
Here $I_H$ stands for indicator function of $H$.
This statement is true for the following borderline cases:
If $X_{1, 1}$ is uniformly distributed on $G$, then $H$ is equal to $G$. This is because if $A$ and $B$ are two independent uniformly distributed random elements of a finite group, then $AB$ is also uniformly distributed.
If $X_{1, 1}$ is degenerate ($\exists g \in G$, such that $P(X_{1, 1} = g) = 1)$), then $H$ is trivial, as $\forall g \in G$ $g^{ord(g)} = e$ by definition of group element order.
But is that statement true in general?
As in my comment, there seems to be a counterexample in the degenerate case: when $G = \mathbb{Z}/m\mathbb{Z}$ and $g = 1$, we see that $A_n$ alternates so that $A_{2n} = 0$ (with probability $1$) and $A_{2n+1} = 1$ (with probability $1$), meaning no such $H$ exists.