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Let $mathcal U$ be a free ultrafilter on a set $X$. For $ninmathbb N$ let $mathcal F$ be a family of $n$-element subsets of $X$ such that $bigcupmathcal Finmathcal U$.

Question. Is there a set $Uinmathcal U$ and a subfamily $mathcal Esubsetmathcal F$ such that $bigcupmathcal Einmathcal U$ and $|Ucap E|=1$ for every $Einmathcal E$?

I do not know the answer even for $n=2$ and countable set $X$.

Here's a try for $n=3$ which I think generalizes to any $n$.

Well-order $mathcalF$ as $F_alpha$. By discarding any $F_alpha$ which is contained in $bigcup_beta <alpha F_beta$, we can ensure that each $F_alpha$ contains at least one point which is not in any previous $F_beta$, without affecting $U = bigcup F_alpha$.

We can now inductively construct a set $V subset U$ with the property that $|V cap F_alpha|= 1$ or $2$ for all alpha. At each $alpha$ at least one element of $F_alpha$ is new and we can include it or not to ensure the condition for $F_alpha$.

Now $U setminus V$ has the same property, so since $U in mathcalU$, wlog we can assume $V in mathcalU$.

Let $mathcalE_1 subseteq mathcalF$ consist of those $F_alpha$ which intersect $V$ in exactly one point, and let $mathcalE_2$ consist of the other $F_alpha$, which all intersect $V$ in two points. If the union of $mathcalE_1$ belongs to $mathcalU$ then we are done. Otherwise the union of $mathcalE_2$ belongs to $mathcalU$ and this reduces us to the $n=2$ case.

The general reduction turns the problem for an arbitrary $n$ into the same problem for some smaller $n$.

For $n=2$ and any $X$: for each $xin cup mathcal F$ fix any edge $(x,y)in mathcal F$ and draw an arrow from $x$ to $y$. Remove all other edges. Remaining edges still have the same union as $mathcalF$, but they form a directed graph with outdegrees at most 1. Such a graph has a proper 3-coloring (it suffices to prove this for finite subgraphs by compactness theorem, and for them the result is well known and easy: remove the vertex with minimal in-degree, it is at most 1, and proceed by induction). One of three colors works.