Does $\text{Aut}(A)\cong \text{Aut}(B)\cong \text{Aut}(A\bigcap B)$ imply $A=B$? The question is in the title: given two mathematical structures $A$ and $B$ that we conjecture to be equal, is it sufficient to prove that $\text{Aut}(A)\cong \text{Aut}(B)\cong \text{Aut}(A\bigcap B)$, where $\text{Aut}(X)$ is the automorphism group of $X$, to establish the desired equality $A=B$?
Thanks in advance.
EDIT December 27th 2013: Stated the way it is, the question can't be answered affirmatively. What if we add as an hypothesis that there exists a measure $\mu$ such that $\mu(A\Delta B)=0$?
 A: Presumably $A$ and $B$ are subsets of a bigger set, because otherwise you cannot speak of their intersection. Assuming this so that the question makes sense, it is false. For instance, consider the abelian groups $A=2\mathbf Z$ and $B=3\mathbf Z$ sitting inside $\mathbf Z$. The intersection is $6 \mathbf Z$. The automorphism groups of all three objects are isomorphic to $\pm 1$, but $A \neq B$.
Edit: Even if you want to relax the conclusion from an equality $A=B$ to an isomorphism $A\cong B$, it is still false. For instance, the automorphism group of the one-point set is isomorphic to the automorphism group of its empty subset (they are both trivial), but the empty set and the one-point set are not isomorphic.
A: A counterexample to the case where we have your condition  but A and B are not even isomorphic.
Consider the fields $\mathbb{R}$ and $\mathbb{Q}$. Then the automorphism groups all consist solely of the identity, but $\mathbb{R}$ is not isomorphic to $\mathbb{Q}$.
A: The answer is no. In fact, there is quite strong reason for it.
Take for example the group (or vector space) $\mathbb{Z}_2^{\omega}$, i.e. the countable product of copies of $\mathbb{Z}_2$. Then the subgroups $A=\{(a_n)_{n<\omega}\;|\; a_0=0\}$ and $B=\{(a_n)_{n<\omega}\;|\; a_1=0\}$ are isomorphic to each other and, moreover, both are isomorphic to their intersection $A\cap B=\{(a_n)_{n<\omega}\;|\; a_0=a_1=0\}$. Then of course, the respective automorphism groups are isomorphic to each other as well, but all of the three sets are distinct.
