For a complete truth-set $T$ is a countable transitive model satisfying $T$ unique? Let $T$ be a maximal (in the sense that either $\phi \in T$ or $\phi \not \in T$ for all $\phi \in \mathcal{L}_\in$) set of sentences consistent with $ZFC$.
Question For a countable transitive model $\mathfrak{M}$ such that $\mathfrak{M} \models T$, is $\mathfrak{M}$ the unique such model?
This is somewhat close to questions asked here:
On the number of countable models of complete theories of models of ZFC
and here:
Number of Non-isomorphic models of Set Theory
However, those answers just deal with countable models rather than countable transitive models. 
It is tempting to say that $\mathfrak{M}$ should be unique; for Skolem hull $H \models T$ the collapsing isomorphism to $\mathfrak{M}$ is unique. It's then hard to see what could substantiate any difference between the result of collapsing two distinct $H_1$ and $H_2$ that both satisfy $T$.
Many thanks for any pointers! 
 A: Sometimes the answer is yes, for example consider $T$ which includes these two axioms:


*

*$V=L$.

*"There are no transitive models of $\sf ZFC$".


Because of the first statement any countable transitive model has the form $L_\alpha$ for some $\alpha<\omega_1$. But the second statement already ensures that this $\alpha$ is unique, since any larger $\alpha'$ will know $L_\alpha$ and therefore the second axiom will be false there.
Similarly we can add all sort of these statements which make the choice of $\alpha$ unique.
Sometimes the answer is no, for example if $0^\#$ exists, then there is a class of indiscernibles for $L$, in particular if $\alpha$ and $\beta$ are indiscernibles then $L_\alpha$ is an elementary submodel of $L_\beta$. Pick $\alpha$ any countable indiscernible (and there are many of them), and take $T$ as the theory of $L_\alpha$, now you have $\aleph_1$ non-isomorphic countable transitive models of $T$.
And sometimes the answer is "what are you talking about?" consider the model from the first example, and work internally to that model. Regardless to what completion we took of $\sf ZFC$ it doesn't have a transitive model, let alone a countable transitive model. 
A: Here's a particularly simple way to show non-uniqueness for arbitrary T. Suppose there's a transitive model $M$ of ZFC containing $\mathcal P(\omega)$. Let T be the complete theory of $M$. Now, suppose there are less than $2^\omega$ ctm's (up to isomorphism) of T. Then there is some $x\subseteq \omega$ not (represented) in any of them. But the countable Skolem hull of $M$ wrt $\{x\}$ will have a transitive collapse containing $x$. Contradiction.
