Do two open diffeomorphic sets in $\mathbb{R^n}$ with the same boundary coincide? Let $A,B \subseteq \mathbb{R}^n$  be open diffeomorphic subsets such that $\partial A=\partial B$. 
Is it true that $A=B$?
(I have seen this claim somewhere but with no proof).
Edit:
As commented by Noah, this is not true in general: e.g take $A,B$ to be the left and right half-planes in $\mathbb{R}^2$.
The question might be more interesting if we require $A,B$ to be bounded.
 A: Lemma. Suppose that $M_1, M_2$ are codimension $0$ compact topological submanifolds of $E^n$ which have common boundary $C$. Then $M_1=M_2$.
Proof. Let $f_i: M_i\to E^n$ be the inclusion maps. Consider a manifold $M$ obtained from $M_1\sqcup M_2$ by gluing the boundaries of these manifolds via  the homeomorphism 
$$h=f_2^{-1}\circ f_1|_{\partial M_1}.$$
Then $M$ is a closed topological $n$-dimensional manifold without boundary. It comes equipped with a continuous map $f: M\to E^n$ as well as inclusion maps $\iota_k: M_k\to M$, $k=1,2$. The maps $g_k=f\circ \iota_k, k=1, 2$ are topological embeddings. I will leave it to you to check that $M$ is oriented so that the map $\iota_1$ preserves the orientation and $\iota_2$ reverses the natural orientation on the manifolds $M_1, M_2$ respectively. 
If there is a point $x\in E^n - C$ such that $f^{-1}(x)$ is a singleton then there is also a small open ball $B\subset M$ containing $x$ such that for every $y\in f(B)$, $f^{-1}(y)$ is a singleton. Hence, we conclude that the map $f: M\to E^n$ has nonzero degree, which is impossible since $E^n$ is contractible. Thus, the restrictions of $g_1, g_2$ to the interiors of $M_1, M_2$ have the same image. In other words, $M_1=M_2$, regarded as subsets of $E^n$.  qed
Edit. For the sake of completeness: There exists a countably infinite collection of bounded open contractible subsets of $E^2$ (all homeomorphic, of course), which all have the same boundary. This example (due to Takeo Wada) is called "Lakes of Wada" (the original example had 3 such subsets). This example was discussed numerous times at MSE.
A: Just to for completeness:
It's not true that $A=B$, and boundedness assumption cannot help, since $\mathbb{R}^n$ is homeomorphic to a bounded subset of itself.
