What are the intermediate subfactors of the tensor product of two maximal subfactors? Let $(N_1 \subset M_1)$ and $(N_2 \subset M_2)$ be two maximal subfactors.    
Their tensor product, the subfactor $(N_1 \otimes N_2 \subset M_1 \otimes M_2)$, admits four obvious intermediate subfactors : $N_1 \otimes N_2$,  $M_1 \otimes M_2$,  $N_1 \otimes M_2$ and $M_1 \otimes N_2$, but what about the non-obvious ?  
For example, the tensor product of $(R^{\mathbb{Z}/p\mathbb{Z}} \subset R)$ and $(R^{\mathbb{Z}/q\mathbb{Z}} \subset R)$, with $p$ and $q$ prime numbers,   admits (at least) one non-obvious intermediate subfactor if and only if $p=q$.  
So in general we could speculate  that there is (at least) one non-obvious intermediate subfactor if and only if the initial subfactors are isomorphic, but it's  false  if they are 2-supertransitive and of index > 2, because in this case, even if they are isomorphic, there is no room for a single one non-obvious.
(see Watatani (1996) prop5.1 p329)  

What are the possible cases about non-obvious for the tensor
  product of two maximal subfactors ?


Remark : the intermediate subfactor lattices $\mathcal{L}(R^{\mathbb{Z}/p\mathbb{Z}} \otimes R^{\mathbb{Z}/p\mathbb{Z}}  \subset R \otimes R)$ are different for each $p$.
Theorem (Lukacs-Palfy 1986): Let $G$ be a finite abelian group, $H$ a group. If the subgroups lattice of $G \times G$ and $H \times H$ are isomorphic then $G \simeq H$.
 A: This answer came after a discussion with Feng Xu. The following more general result is true:  
Theorem:
Let $(N_i \subset M_i)$, $i=1,2$, be irreducible finite index subfactors. Then $$  \mathcal{L}(N_1  \subset M_1) \times \mathcal{L}(N_2 \subset  M_2) \subsetneq \mathcal{L}(N_1 \otimes N_2 \subset M_1 \otimes M_2)$$ if and only if they are intermediate subfactors $N_i \subseteq P_i \subset Q_i \subseteq M_i$,  $i=1,2$, such that $(P_i \subset Q_i)$ is depth $2$ and isomorphic to $(R^{\mathbb{A}_i} \subset R)$, with $\mathbb{A}_2 \simeq \mathbb{A}_1^{cop}$ which is the Kac algebra $\mathbb{A}_1$ with the opposite coproduct.  
Proof:
This theorem was proved  in the $2$-supertransitive case by Y. Watatani. The general case was conjectured by the OP, and proved by a discussion with Feng Xu as follows:
Let the intermediate subfactors $$N_1 \otimes N_2 \subseteq P_1 \otimes P_2 \subset R \subset Q_1 \otimes Q_2 \subseteq M_1 \otimes M_2$$ with $R$ not of tensor product form, $P_1 \otimes P_2$ and $Q_1 \otimes Q_2$ the closest (below and above resp.) to $R$ among them of tensor product form. Now using Proposition 3.5 (2) of [xu], $(P_i \subseteq Q_i)$, $i=1,2$,  are depth $2$, there corresponding Kac algebras, $\mathbb{A}_i$, $i=1,2$, are very simple and $\mathbb{A}_2 \simeq \mathbb{A}_1^{cop}$ (see Definition 3.6 and Proposition 3.10 of [xu]). The converse is given by Theorem 3.14 (2) of [xu].
A: 
Partial answer : for the group-subgroup subfactors $(R^G \subset R^H)$  

Theorem:  Let $(H_i \subset G_i)$ be core-free maximal inclusions of groups, then $(H_1 \times H_2 \subset G_1 \times G_2)$ admits a non-obvious intermediate subgroup iff $G_1 \simeq G_2 \simeq \mathbb{Z}_p$.
Proof : see this answer. $\square$    

Corollary: The tensor product of two group-subgroup maximal subfactors admits a non-obvious intermediate iff the
  subfactors are isomorphic and depth $2$.

Proof : if $K \subset H$ is a normal subgroup of $G$ then $(R^G \subset R^H) \simeq (R^{G/K} \subset R^{H/K})$.
So we can restrict to the group-subgroup subfactors $(R^G \subset R^H)$ with $H$ a core-free subgroup of $G$.
But $(R^{G_1} \otimes R^{G_2} \subset R^{H_1} \otimes R^{H_2}) \simeq (R^{G_1 \times G_2} \subset R^{H_1 \times H_2})$, and $(R^G \subset R^H)$ is maximal iff $(H \subset G)$ is maximal, by the Galois correspondence, which proves the result by the theorem. $\square$   

Problem : Is the corollary true for all the (irreducible) maximal subfactors ?    
Remark : It's also true for the $2$-supertransitive subfactors thanks to the result of Watatani cited above.
Now what's about if at least one of them is not $2$-supertransitive ?
And what's about $(R^{G_1} \otimes R\rtimes{H_2} \subset R^{H_1} \otimes R \rtimes {G_2}) $  ?     
A: 
Direct proof (to be completed)  generalizing this argument of groups theory :  

Let $(N_i \subset M_i)$ be irreducible maximal subfactors.
Let $P$ an intermediate subfactor:  $N_1 \otimes N_2  \subset P \subset M_1 \otimes M_2$. 
Let $P^1 = \{ x_1 \in M_1 \text{ such that } \exists x_2 \in M_2 \text{ with } x_1 \otimes x_2 \in P \}''$,
and $P_1 = \{ x_1 \in M_1  \text{ such that } x_1 \otimes N_2 \subset P \}''$
Idem, we define $P^2$ and $P_2$.    
($\star$) To be proved:  $(P_i \subset P^i)$ is depth $2$.  
Then $(P_i \subset P^i) \simeq (R \subset R \rtimes \mathbb{A}_i)$  with $  \mathbb{A}_i$ a Kac algebra.
Let a $W^*$-isomorphism $\psi_i : R \rtimes \mathbb{A}_i \to P^i $  with $\psi_i(R) = P_i $.
Let $\phi : \mathbb{A}_1 \to \mathbb{A}_2$ with $\phi(a_1) = a_2$ such that $\psi_1(a_1) \otimes \psi_2(a_2) \in P$  
($\star$) To be proved: $\phi$ is a well-defined isomorphism of Kac algebras.
Then $(P_1 \subset P^1) \simeq  (P_2 \subset P^2)$. 
Now $N_i \subset P_i \subset P^i \subset M_i$, so by maximality: $P_i, P^i \in \{N_i , M_i   \}$.   
If $(N_1 \subset M_1)$ is depth $>2$ then $P_1=P^1=N_1$ or $M_1$, because $(P_1 \subset P^1)$ is depth $2$,
and $P_2=P^2=N_2$ or $M_2$ because $(P_1 \subset P^1) \simeq (P_2 \subset P^2)$.  
($\star$) To be proved:    $P = \{  x_1 \otimes x_2  \in P \text{ such that } x_1 \in M_1 \text{ and } x_2 \in M_2  \}''$
But $x_1 \otimes x_2 \in P$ implies $x_1 \otimes N_2$, $N_1 \otimes x_2 \subset P$, because $P_i=P^i$.
So $P = P_1 \otimes P_2 \in \{N_1 \otimes N_2 , N_1 \otimes M_2 , M_1 \otimes N_2  , M_1 \otimes M_2    \} $  
If $(N_2 \subset M_2)$ is depth $>2$, idem...   
If  $(N_1 \subset M_1)$ and $ (N_2 \subset M_2)$ are depth $2$, but are not isomorphic, idem...
Else $(N_1 \subset M_1) \simeq (N_2 \subset M_2) \simeq (R \subset R \rtimes \mathbb{A})$ with $\mathbb{A}$ a maximal Kac algebra.
Let $a \not\in \mathbb{C}1$ and $\langle a \rangle$ the left coideal generated by $a$, then by maximalilty  $\langle a \rangle = \mathbb{A}$.
So $\mathbb{A} \simeq \langle a \otimes a \rangle \not\in \{ \mathbb{C}\otimes  \mathbb{C}, \mathbb{C}\otimes  \mathbb{A} , \mathbb{A}\otimes  \mathbb{C} , \mathbb{A}\otimes  \mathbb{A}  \} $, i.e. a non-obvious left coideal of $\mathbb{A}\otimes  \mathbb{A}$.
The result follows by Galois correspondence. $\square$  

Corollary of the proof : the lattice of intermediate subfactors of the tensor product of finitely many irreducible subfactors $(N_i \subset M_i)_{i}$ is the direct product of the lattices $\mathcal{L}(N_i \subset M_i)$ of each subfactors if and only if they are pairwise without isomorphic maximal intermediate depth $2$ inclusions.
Proof : the result follows by using $(P_i \subset P^i)$ as above and induction. 
Corollary : the tensor product of finitely many irreducible cyclic subfactors is a cyclic subfactor if and only if they are pairwise without isomorphic maximal intermediate depth $2$  inclusions.
Proof : the direct product of distributive lattices is also distributive.
Remark: In particular, the tensor product of finitely many irreducible maximal subfactors is a cyclic subfactor if and only if they don't contain two  isomorphic subfactors of depth  $2$. 
