Let $E$ be a Euclidean space, and let $X, Y\subseteq E$ be two disjoint subsets.
Suppose that there exist isometries $i$ and $j$ such that $i(X)\cap j(Y)=\emptyset$ and $i(X)\cup j(Y)=X\cup Y$. Does it follow that one of the following is true: (i) there's an isometry fixing $X\cup Y$, mapping $X$ to $i(X)$ and mapping $Y$ to $j(Y)$, or (ii) there's an isometry fixing $X\cup Y$, mapping $X$ to $j(Y)$ and mapping $Y$ to $i(X)$?
No. Consider two interlocking puzzle pieces, not isometric to each other, and where their union doesn't have any non-trivial symmetries. Here's a picture of the idea.
The only isometry that fixes $X \cup Y$ is the identity map, which does not take $X$ to either $i(X)$ or $j(Y)$.