In ZFC (the axiom of foundation is most essential), a graph is a transitive closure graph if and only if it is extensional (distinct vertices have distinct sets of incoming edges), well-founded (there is no descending infinite path) and there is a unique sink reachable by a path from every vertex. (The left-to-right direction is obvious, for the right-to-left direction the set can be constructed by well-founded recursion.) Moreover, the graph uniquely determines the set. (This can be proved easily by well-founded induction.) Finally, the graph is finite if and only if the corresponding set is hereditarily finite.
For your question, this entails:
$Y$ is unique if it exists.
For finite graphs, well-foundedness is equivalent to there being no directed cycles, which is a symmetric condition, and therefore is automatically satisfied for the converse of $G(X)$. Similarly, $G(X)$ always has a unique source (namely, the vertex corresponding to the empty set) and every vertex is reachable from it by a directed path. Thus,
$Y$ exists if and only if the converse of $G(X)$ is extensional,
which amounts to the following condition:
For every $a\ne b$ in $\operatorname{TC}(\{X\})$, there is $c\in\operatorname{TC}(\{X\})$ such that $a\in c$ and $b\notin c$ or vice versa.
Linda’s example shows that there are sets $X$ failing this condition.
With regards to the “further question”: there is almost no correlation, except for the obvious bound that $|X|,|Y| < |\operatorname{TC}(\{X\})|$. You gave the extreme examples yourself: on the one hand, von Neumann ordinals $n$ have $|X|=|Y|=n$, $|\operatorname{TC}(\{X\})|=n+1$. On the other hand, Zermelo ordinals $n'$ have $|X|=|Y|=1$, $|\operatorname{TC}(\{X\})|=n+1$. For a mixed example, $X=\{1',\dots,n'\}$ has $|X|=n$, $|Y|=1$, $|\operatorname{TC}(\{X\})|=n+2$. It’s easy to cook up similar examples for other combinations of cardinalities.
