Let $X$ be a nonempty finite set. Without loss of generality, we can take $X = \{1, 2 \cdots ,n\}$.
A $\textit{partition}$ $\mathcal{P} = \{X_1, X_2 \cdots X_k\}$ of $X$ is a collection of pairwise disjoint nonempty subsets of $X$ whose union is $X$.
We say a partition $\mathcal{P} = \{X_1, X_2 \cdots X_k\}$ is $\textit{balanced}$ if there exists a positive integer $\ell$ such that $\ell = |X_i|$ for all indices $i$. This number $\ell$ is then called the $\textit{proportion}$ of $\mathcal{P}$ and we have the obvious relation $|X| = \ell k$.
Finally, we say a set $T \subset X$ is a $\textit{traversal}$ of a partition $\mathcal{P} = \{X_1, X_2 \cdots X_k\}$ if it is the image of some function $f:\mathcal{P} \to X$ with the property $f(x) \in x$ for any $x \in \mathcal{P}$.
Conjecture: If $\mathcal{P}_1, \mathcal{P}_2$ are any two balanced partitions of $X$ with the same proportion (say $\ell$), there exists a $T \subset X$ which is a traversal for both $\mathcal{P}_1$ and $\mathcal{P}_2$ (i.e., a simultaneous traversal).
Examples: $X=\{1,2,3,4,5,6\}$ and partitions $\mathcal{P}_1 = \{\{1,2\}, \{3,4\}, \{5,6\}\}$, and $\mathcal{P}_2 = \{\{1,6\}, \{3,5\}, \{4,2\}\}$. Then $T=\{1,4,5\}$ is a simultaneous traversal.
Context: This arises as a generalization of a group theory problem. Specifically, if one has a group $G$ and a (not necessarily normal) subgroup $H$, then one can find a set $T \subset G$ which is a set of representatives for both the left and right cosets of $H$. The proof for this specific claim which I'm aware of uses group-theoretical concepts, so I'm not sure if it can be generalized to solve the above conjecture. Perhaps, in solving the more general problem, Hall's marriage theorem might be useful, although I have not been able to get this to work.
Best Answer
The existence of a simultaneous traversal is equivalent to the following statement:
To see the equivalence, just note that if we had such a bijection, we could choose elements of each set $X_i\cap f(X_i)$ to get a simultaneous traversal. Conversely, if we had a simultaneous traversal $T$, we can set $f(X_i)$ to be the element of $P_2$ containing the singleton $X_i\cap T$.
This statement is straightforwards to prove by Hall's Marriage Theorem. In particular, it becomes equivalent to the following marriage condition:
This is clear because $\bigcup S$ has $|S|\cdot \ell$ elements, so the elements of $\mathcal P_2$ intersecting $\bigcup S$ must cover at least $|S|\cdot \ell$ elements as well, so there must be at least $\ell$ of them.