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 there's an isometry fixing $X\cup Y$, mapping $X$ to $i(X)$ and mapping $Y$ to $j(Y)$? (Hints and pointers welcome too.)
No.
Let $X$ be a tetrahedron such that the only isometry fixing $X$ is the identity. (For example, $X$ could have six different edge lengths.) Now let $i$ be some translation that moves $X$ far away, so that $i(X) \cap X = \emptyset$. Write $Y = i(X)$ and $j = i^{-1}$.
Then $X$ and $Y$ satisfy the conditions. However, $i$ is the only isometry mapping $X$ to $i(X)$, and $i(Y) \ne X = j(Y)$.