Reconstructing a non-$\omega$-categorical countable structure from its automorphism group Let $M$ and $N$ be countable structures.  If $M$ and $N$ are both $\omega$-categorical, then we know that $M$ and $N$ are bi-interpretable iff $\mathrm{Aut}(M) \cong \mathrm{Aut}(N)$ as topological groups.
Suppose that $M$ is $\omega$-categorical and that $N$ is not.  Then:


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*Could it be that $\mathrm{Aut}(M) \cong \mathrm{Aut}(N)$ as topological groups?

*If $\mathrm{Aut}(M) \cong \mathrm{Aut}(N)$ as topological groups, what can be said about the interpretability of $M$ in $N$?  (I'm guessing not much.)

 A: In the comments, you got an example of two countable structures with isomorphic automorphism groups which aren't bi-interpretable. But neither of these structures were $\omega$-categorical. Here's another (silly) example answering your question 1. 
Let $M$ be a countable set in the empty language. Let $L$ be the language with countably many constant symbols, and let $N$ be the countable $L$-structure in which the constant symbols are interpreted as distinct elements, and there are also infinitely many elements not named by constant symbols. Then $\text{Aut}(M)\cong \text{Aut}(N) \cong S_\infty$, the full permutation group of $\omega$. $M$ is $\omega$-categorical, but $N$ is not (the countable models of $\text{Th}(N)$ consist of the constants together with a number of extra elements from $\{0,1,2,\dots\}\cup \{\aleph_0\}$). 
Of course, here $M$ is still interpretable in $N$. To get a counterexample to question 2, we have to work a bit harder. 
This MathOverflow answer shows that there is a predicate $A$ on $\mathbb{Z}$ such that $(\mathbb{Z},<,A)$ has only the trivial automorphism, but this structure also has no first-order definable elements (indeed, you can pick $A$ at random, and it works with probability $1$!). 
Now let $M = (\mathbb{Q},<)$, and consider the structure $N = (\mathbb{Q}\times \mathbb{Z}, <, P)$, where $<$ is the lexicographic order on $\mathbb{Q}\times \mathbb{Z}$, and $P$ is a unary prediate consisting of $A$ on each copy of $\mathbb{Z}$, i.e. $P = \{(q,n)\mid n\in A\}$. 
Any automorphism of $N$ permutes the copies of $\mathbb{Z}$. And since $(\mathbb{Z},<,A)$ is rigid, there is exactly one isomorphism between any two copies of $\mathbb{Z}$. So $\text{Aut}(N)\cong \text{Aut}(M)$. 
Now it should be possible to prove that $M$ is not interpretable in $N$, by showing that $N$ does not admit any non-trivial definable equivalence relations, and the copies of $\mathbb{Z}$ in $N$ contain no definable elements. The point is that the natural way to interpret $(\mathbb{Q},<)$ would be in the quotient by the equivalence relation $x\sim y$ iff there are only finitely many elements between $x$ and $y$ (here an equivalence class is a copy of $\mathbb{Z}$), or to take the domain to consist of a single representative from each equivalence class. But this equivalence relation is not definable, and it has no definable set of representatives. 
