Let $U$ be a set.
Suppose $f:\mathscr{P}U\to\mathscr{P}U$ is a function satisfying $f(\varnothing)=\varnothing$ and $f(A\cup B)=f(A)\cup f(B)$.
I denote $\left(\bigsqcup_{f\in S}f\right)(X) = \bigcup_{f\in S}f(X)$.
Conjecture $(1\sqcup f\sqcup f^2\sqcup\dots)\circ(1\sqcup f\sqcup f^2\sqcup\dots) = 1\sqcup f\sqcup f^2\sqcup\dots$
(here $1$ is the identity relation on $U$ and $\circ$ is composition of functions with $f^n = f\circ\dots\circ f$ ($n$ times)).
Counterexample (please check for errors):
Consider $f:\mathscr{P}\mathbb{R}\to\mathscr{P}\mathbb{R}$ defined by the formula
$$ f (E) = \operatorname{cl} \bigcup \left\{ \left[ \frac{- 1 - | a |}{2} ; \frac{1 + | a |}{2} \right] \mid a \in E \right\} $$
(here $\operatorname{cl}$ is the usual topological closure for real numbers).
Take $X = \left[ - \frac{1}{2} ; \frac{1}{2} \right]$. $f^n (X) = \left[ - 1 + \frac{1}{2^{n+1}} ; 1 - \frac{1}{2^{n+1}} \right]$.
We have $(1 \sqcup f \sqcup f^2 \sqcup \ldots ) (X) =] - 1 ; 1 [$;
$(1 \sqcup f \sqcup f^2 \sqcup \ldots )((1 \sqcup f \sqcup f^2 \sqcup \ldots) (X)) = [- 1 ; 1]$.
Thus follows $(1 \sqcup f \sqcup f^2 \sqcup \ldots) \circ (1 \sqcup f \sqcup f^2 \sqcup \ldots) \neq 1 \sqcup f \sqcup f^2 \sqcup ...$