Let $T$ be a theory and let $I$ be an imaginary sort of $T$. We'll let $T_I$ denote the induced theory on $I$, by which I mean specifically the theory of all $\varnothing$-definable relations on $I$.
Suppose that
for all models $\mathfrak{I} \models T_I$ there is an $\mathfrak{M} \models T$ such that $\mathfrak{I} \cong I(\mathfrak{M})$ and furthermore $\mathfrak{M}$ is unique in the sense that for any two $\mathfrak{M}_0$ and $\mathfrak{M}_1$ with isomorphisms $f_i : I(\mathfrak{M}_i) \cong \mathfrak{I}$ there is an isomorphism $g:\mathfrak{M}_0 \cong \mathfrak{M}_1$ such that $$f_1^{-1} \circ I(g) \circ f_0 = \mathrm{id}_\mathfrak{I},$$ where $I(g): I(\mathfrak{M}_0)\cong I(\mathfrak{M}_1)$ is the induced isomorphism and $\mathrm{id}_{\mathfrak{I}}$ is the identity on $\mathfrak{I}$, and
the home sort of $T$ is 'extensional' over $I$ in the sense that for any $a,b \in \mathfrak{M} \models T$, we have that $a=b$ if and only if $a\equiv_{I(\mathfrak{M})} b$.
Does it follow that the interpretation of $T_I$ in $T$ extends to a bi-interpretation between $T$ and $T_I$?
The conclusion does not follow from either 1 or 2 alone:
- Let $T$ be the theory of an equivalence relation with infinitely many equivalence classes of size $2$, and let $I$ be the imaginary that results from modding by the equivalence relation. $T$ and $I$ satisfy 1 but not 2 and fail the conclusion.
- Let $T$ be the theory of the collection of subsets of $\omega$ with $\subseteq$. The set of singletons is definable, so we can define an equivalence relation where $xEy$ if either $x$ and $y$ are both singletons and $x=y$ or $x$ and $y$ are both not singletons. If we let $I$ be the imaginary resulting from modding by this equivalence relation then $T$ and $I$ satisfy 2 but not 1 and fail the conclusion.