Derived equivalence commutes with standard duality Let $A$ and $B$ be derived equivalent $k$-algebras, for a field $k$. Denote $DA$ the derived category of $modA$, the module category of $A$. Let $F:DA \to DB$ be a triangule equivalence. It is well known that $F$ can be extended to an equivalence $$\tilde{F}: DA^e \to DB^e$$ where $A^e=A \otimes_k A^{op}$ is the enveloping algebra of $A$. Let $*:modA \to modA$ be the standard duality, i.e $N^* = Hom_k(N,k)$. This functor extends to a functor $*:DA^e \to DA^e$, and as well for $DB^e$.
My question is, given an $A$-bimodule $N$ suppose $\tilde{F}(N)$ is concentrated in degree $0$ so it can be considered as an $A$-bimodule, is it true that $\tilde{F}(N^*) \cong \tilde{F}(N)^*$?
 A: In fact, more is true. Suppose $(A,B,X,Y)$ are the data for a derived equivalence, where $A$ and $B$ are $k$-algebras, $X$ is a two-sided tilting complex in $D(A^{op}\otimes_kB)$, so $-\otimes^{\bf L}_AX$ is an equivalence from the derived category $D(A)$ to $D(B)$ (my modules are right modules), and $Y$ is the two-sided tilting complex in $D(B^{op}\otimes_kA)$ inducing the adjoint equivalence. Then for $N$ in $D(A)$, 
$$(N\otimes^{\bf L}_AX)^\ast\cong Y\otimes^{\bf L}_AN^\ast.$$
This answers your question if you take the data $(A^e,B^e,X\otimes_kY,Y\otimes_kX)$, and your assumptions that $N$ and $\tilde{F}(N)$ are concentrated in degree zero aren't needed.
To prove my claim, consider the functor
$$D(B^{op})\to D(k)$$
given by 
$$M\mapsto N\otimes^{\bf L}_AX\otimes^{\bf L}_BM.$$ 
This has right adjoint
$$V\mapsto\text{Hom}_k(N\otimes^{\bf L}_AX,V),$$
which takes the value $(N\otimes^{\bf L}_AX)^\ast$ on $V=k$.
But also it is the composition of functors
$$D(B^{op})\to D(A^{op})\to D(k)$$
given by $M\mapsto X\otimes^{\bf L}_BM$ and $L\mapsto N\otimes^{\bf L}_AL$, and so it also has right adjoint
$$V\mapsto Y\otimes^{\bf L}_A\text{Hom}_k(N,V),$$
which takes the value $Y\otimes^{\bf L}_AN^\ast$ on $V=k$.
So $(N\otimes^{\bf L}_AX)^\ast\cong Y\otimes^{\bf L}_AN^\ast$.
