An interpretation of one theory $T$ in another theory $S$ is essentially an appropriate tuple of formulas $\Phi$ (one formula for each symbol in the language of $S$, plus another formula for "domain") such that in any model $M$ of $T$, we have "$\Phi^T\models S$" - that is, $\Phi$ defines in $M$ a set and some relations/functions on that set which together form a model of $S$.
For example, consider the usual construction of $\mathbb{C}$ as ordered pairs of reals. This corresponds to an interpretation of $Th(\mathbb{C};+,\cdot)$ in $Th(\mathbb{R};+,\cdot)$; the formulas in question are
$\varphi_{domain}(x,y)\equiv \top$ (the domain is all ordered pairs),
$\varphi_+(x_1,x_2,y_1,y_2,z_1,z_2)\equiv (x_1+y_1=z_1)\wedge (x_2+y_2=z_2)$ (addition is componentwise), and
$\varphi_\cdot(x_1,x_2,y_1,y_2,z_1,z_2)\equiv (x_1\cdot y_1-x_2\cdot y_2=z_1)\wedge(x_1\cdot y_2+x_2\cdot y_1=z_2)$ (multiplication treats the second coordinate like $i$ - note that I'm using subtraction here, which I shouldn't really, but since it's definable in terms of $+$ this is a benign abbreviation).
(A precise definition can be found in either of Hodges' model theory books, and this blog post of Hamkins - which I also cite below for a technical result - gives a good description of the notion.)
For a more technical example, forcing gives a method for constructing interpretations: e.g. the proof that ZFC+$\neg$CH is consistent if ZFC is ultimately amounts to constructing an interpretation $\Phi$ of ZFC+$\neg$CH in ZFC (with a bit of thought taken to handle non-well-founded models).
Two theories are then inter-interpretable (or mutually interpretable) if each is interpretable in the other. Interestingly there is a sharper version, called bi-interpretability, which requires that the interpretations commute nicely: given $M\models T$, we need "($M$'s model of $S$)'s model of $T$" to be isomorphic to $M$ (and similarly with $S$ and $T$ switched).
(In fact we require even more than that even: we need such isomorphisms to be uniformly definable. But this is a bit technical so I'll ignore it for now.)
This is in general a stronger condition than mere mutual interpretability; see e.g. here to see a stark difference.
Best Answer
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.