"Composition of Morita equivalences" or "Morita equivalence and the Nakayama functor" This problem occured to me, when trying to find a Morita invariant for finite dimensional algebras.
Suppose $\Lambda$ and $\Gamma$ are two self-injective $k$-algebras ($k$ being a field) which are Morita equivalent. The Morita equivalence should be given by the $\Lambda-\Gamma$-bimodule $P$ and the $\Gamma$-$\Lambda$-bimodule $Q$. Each of the algebras has a Morita self-equivalence given by the bimodules $D\Lambda$ and $D\Gamma$, where $D$ is the standard $k$-duality. 

What can one say about the
  commutativity of the two compositions,
  i.e. is $$D\Gamma\otimes_{\Gamma}
Q\otimes_\Lambda - \cong
Q\otimes_\Lambda D\Lambda
\otimes_\Lambda - $$ in general?

Further comment: The Morita self-equivalence can equivalently be given as $D \operatorname{Hom}_\Lambda(-,\Lambda)$. This is called the Nakayama functor.
I did not succeed in finding a counterexample nor giving a proof and I don't know what to expect.
 A: Proposition 5.2 in J. Rickard's paper "Derived Equivalences as Derived Functors" seems to show what you want to show, however for standard derived equivalences and the left derived Nakayama functors. I.e., Rickard claims $$D\Gamma \otimes_\Gamma^{\mathbb L}Q \otimes_\Lambda^{\mathbb L}- \cong Q \otimes_\Lambda^{\mathbb L} D\Lambda\otimes_\Lambda^{\mathbb L}-$$
as functors from $D^-(\textrm{Mod-}\Lambda)$ to $D^-(\textrm{Mod-}\Gamma)$. Your claim is, I believe, a corollary of this. The reason for that is that $D\Gamma \otimes_\Gamma^{\mathbb L}Q \otimes_\Lambda^{\mathbb L}-$ is isomorphic to $(D\Gamma \otimes_\Gamma Q) \otimes_\Lambda^{\mathbb L}-$ (since $Q$ is projective as a left $\Gamma$-module) and 
$Q \otimes_\Lambda^{\mathbb L} D\Lambda\otimes_\Lambda^{\mathbb L}- \cong (Q \otimes_\Lambda D\Lambda)\otimes_\Lambda^{\mathbb L}-$, and I suppose this implies $D\Gamma \otimes_\Gamma Q\cong Q\otimes_\Lambda D\Lambda$ as $\Gamma$-$\Lambda$-bimodules.  
A: I'm sure you've already considered this, but it is trivially true if $\Lambda$ and $\Gamma$ are symmetric (i.e. group algebras), since we then have isomorphisms $D\Gamma \cong \Gamma$ and $D\Lambda\cong \Lambda$.
I will have a bit more of a think, but I'd bet it does hold for self-injective algebras in general.  
