Example for non equivalent rational full CFTs with same modular invariant (partition function) I am looking for a counter example which shows, that a full rational 2D CFT (with respect to a given chiral subtheory) is not characterized by its modular invariant partition function. People tell me such an example exists, but noone could point out a concrete example or reference to me.

In the framework of Fuchs, Runkel, Schweigert etc. full CFTs are classified by Morita equivalence classes of Frobenius algebras in a modular tensor category $\mathcal C$ (the representation category of the chiral CFT), so my question more concretely:

I am looking for two special symmetric haploid Frobenius algebra objects $A,A'$ in a modular tensor category $\mathcal C$, which are not Morita equivalent but whose full centres $Z(A),Z(A')$ are equivalent as objects (they cannot be equivalent as algebra objects due to a result of Kong and Runkel), i.e. have the same modular invariant partition function $Z(A)_{ij}=Z(A')_{ij}$.


I am also fine with an example in any other formulation of CFT any reference in physics literature, non-degenerate braided subfactors or similar...
 A: There is an SU(3) example here in arXiv:math/0008056 pages 21 and 22 for $SU(3)_9$ in $E_6$ and also in $E_6 \times  Z_3$ - or see also pages 11 and 12 of arXiv:1002.2348,  page 6 of arXiv:0906.4252
A: I'm going to try to connect the language of Jeff Harvey's answer to the language of your question.  It seems to give an answer to your title question, but not to the question you asked in the main text.
If you are given an even unimodular lattice $L$, the resulting vertex algebra $V_L$ has trivial module category, i.e., equivalent to complex vector spaces.  From this datum, you can make a CFT in the Fuchs-Runkel-Schweigert sense by taking the trivial unital algebra in this category, analogous to taking the algebra $\mathbb{C}$ in the category of $\mathbb{C}$-vector spaces.  The partition function is equal to $\frac{\theta_L}{\eta^{rank(L)}}$, which is modular-invariant up to an order 3 character.
If you have two even unimodular lattices, then you get two equivalent categories (both equivalent to the category of complex vector spaces), and if you take the trivial algebras, then the full CFTs are considered Morita equivalent in this definition.  In other words, I suspect this is not an answer to the precise question you asked.  However, in the case of the two rank 16 examples with equal theta functions (hence equal partition functions), there is a very meaningful sense in which the CFTs are not equivalent, since the vertex algebras have nonisomorphic automorphism groups.  In particular, the CFTs have different sets of twining characters, and the corresponding orbifold theories are also quite different.
Regarding the question of intersections in the comments, it follows from the $I\!I_{m,n}$ construction (see Wikipedia) that one can choose embeddings in Euclidean space so that the intersection of lattices is finite index, hence full rank.  In particular, both $I\!I_{8,0} \times I\!I_{8,0}$ and $I\!I_{16,0}$ contain $(2\mathbb{Z})^{\oplus 16}$.
A: I apologize since my answer will involve shameless self-promotion. You can find one example
of this kind on page 36 of my slides. In this example one Frobenius algebra is commutative
and another is not Morita equivalent to a commutative algebra.
A: I have no idea what a ``symmetric haploid Frobenius algebra" is, so my answer may be terribly naive, but given an even-self dual lattice of rank $d$ there is a well known lattice construction of a rational (actually holomorphic) conformal field theory with central charge $c=d$. There are two even self-dual lattices of rank $16$ ($E_8 \times E_8$ and $Spin(32)/(Z/2)$) and these give rise to two inequivalent holomorphic Conformal Field Theories with $c=16$. The partition function of these theories is $\Theta_{\Gamma}/\eta^{16}$ where $\eta$ is the Dedekind eta function and $\Theta_\Gamma$ is the theta function of the associated lattice. The theta functions for these two lattices are the same because they are modular forms of weight $8$ and there is a unique such modular form. This is related to the famous "can one hear the shape of a drum" problem. Thus this pair of lattices gives an example of two inequivalent CFT's with the same partition function.
A: [Edit: my answer is wrong.]
The set of full CFTs with a given modular invariant is cassified by a kind of second cohomology that was introduced in the paper "On a subfactor analogue of the second cohomology" by
Izumi and Kosaki.
This "second cohomology" $H^2(N\subset M)$ is a pointed set that classifies those extensions $N\subset \tilde M$ with the property that ${}_NL^2M_N\cong {}_NL^2\tilde M_N$ (isomorphism of $N$-$N$-bimodules).
In that same paper, the authors show that when the subfactor is of the form $M^G\subset M$ for the outer action of a finite group $G$, then their $H^2$ is equal to the 2nd group cohomolgoy with $U(1)$ coefficients (also equal to $H^3(BG,\mathbb{Z})$).
Now take a conformal net $\mathcal A$, take a finite group $G$ that acts on it, and
consider the inclusion $\mathcal A^G\subset \mathcal A$.
Now, you probably know that full CFTs with left and right chiral algebras given by $\mathcal A^G$
correspond bijectively to relatively local extensions of $\mathcal A^G\subset \mathcal B$, which correspond bijectively to $Q$-systems in $Rep(\mathcal A^G)$ (a.k.a. Frobenius algebra objects).
The modular invariant remembers only the isomorphism class of the vacuum Hilbert space of $\mathcal B$ as an object of $Rep(\mathcal A^G)$, which is the same as remembering the isomorphism class of the bimodule $\{\}_ \{\mathcal A^G(I)\} L^2\mathcal B(I)_ \{\mathcal A^G(I)\}$.
The extension $\mathcal A^G\subset \mathcal A$ corresponds to a given modular invariant of $\mathcal A^G$, and there will be as many other extensions $\mathcal A^G\subset \mathcal B$ with same modular invariant 
as there are elements in the group $H^2(G,U(1))$.
