Is it true that $X|Z_1 \sim Y|Z_2 \iff X = f(Z_1) + \epsilon_x,Y = f(Z_2) + \epsilon_y$ where $\epsilon_x|Z_1\sim\epsilon_y|Z_2$, for some $f(\cdot)$

Question

Is the following claim true?

$\textbf{Claim}$

1. Suppose random variables $Z_1, Z_2$ have a common support $\mathcal{Z}$. Then if there exists a function $f(\cdot)$ and for any $s \in \mathcal{Z}$ a distribution $F(\cdot|s)$ such that $$X = f(Z_1) + \epsilon_x$$ $$Y = f(Z_2) + \epsilon_y$$ where $\{\epsilon_x|Z_1=s\} \sim F(\cdot|s), \{\epsilon_y|Z_2=s\} \sim F(\cdot|s)$; Then $X|Z_1 \sim Y|Z_2$ (where $\sim$ means "has the same distribution").

2. If $X|Z_1 \sim Y|Z_2$, and $\mathbb{E}[X|Z_1], \mathbb{E}[Y|Z_2]$ are finite, then there exist a function $f(\cdot)$, and for any $s \in \mathcal{Z}$ a distribution $F(\cdot|s)$ such that $$X = f(Z_1) + \epsilon_x$$ $$Y = f(Z_2) + \epsilon_y$$ where $\mathbb{E}[\epsilon_x|Z_1] = \mathbb{E}[\epsilon_y|Z_2] = 0$, and $\{\epsilon_x|Z_1=s\} \sim \{\epsilon_y|Z_2=s \}\sim F(\cdot|s)$;

$\textbf{Proof}$

1: Obvious.

2: $$\forall\ s \in \mathcal{Z}:\ X = X + \mathbb{E}[X|Z_1 = s] - \mathbb{E}[X|Z_1 = s] = \mathbb{E}[X|Z_1 = s] + (X - \mathbb{E}[X|Z_1 = s])$$ Similarly, $$Y = \mathbb{E}[Y|Z_2 = s] + (Y - \mathbb{E}[Y|Z_2 = s])$$ Since $X|Z_1 \sim Y|Z_2 \implies \mathbb{E}[X|Z_1 = s] = \mathbb{E}[Y|Z_2 = s]$. Thus, define $$f(s) = \mathbb{E}[X|Z_1 = s] = \mathbb{E}[Y|Z_2 = s]. $$ Furthermore, define $$\epsilon_x = X - \mathbb{E}[X|Z_1],\ \epsilon_y = Y - \mathbb{E}[Y|Z_2].$$ Clearly, $\mathbb{E}[\epsilon_x|Z_1] = \mathbb{E}[\epsilon_y|Z_2] = 0$; Additionally, from the fact that $X|Z_1 \sim Y|Z_2$ it follows that $$\epsilon_x|Z_1 \sim \epsilon_y|Z_2$$

$\textbf{End of the proof}$

If arguments above are correct, is this a some known result/special case of some general result? Would be super grateful for any references! Thanks!