Let $B$ be a $A$-algebra which is free of finite rank as $A$-module. Let $X$ be a finitely generated projective left $B$ module. (So $X$ is also a f.g. projective $A$ module.) Are these homomorphism groups naturally isomorphic as right $B$-modules? $$\mathrm{Hom}_A(X,A)\stackrel{?}{\cong} \mathrm{Hom}_B(X,B).$$

By dimension count (e.g. in special case of fields) one can see easily that these two are really isomorphic. After a lot of work I found a long proof of natural equivalence which only works for the case of division rings with an assumption on characteristic, which doesn't give an explicit isomorphism.

So my question is: Are these two $B$-modules always naturally isomorphic? (At least for the case of a division ring $B$ over a field $A$)

I also wonder if there is a simple explicit natural morphism in general (without above restrictions on $B$ and $X$ ) from one side to the other which gives a natural isomorphism in good cases.

**Edit**. I simplified my proof and found an explicit natural morphism from the right side to the left. Let $f \in \mathrm{Hom}_B(X,B)$, I define the corresponding element $t_f\in \mathrm{Hom}_A(X,A)$ as follows: For every $x\in X$ define a $A$-linear morphism $T_f(x):X\to X$ by $(T_f(x))(y) := f(y)x$. Then $t_f(x):= \mathrm{tr}_A(T_f(x)).$
I can show that $t:\mathrm{Hom}_B(X,B)\to \mathrm{Hom}_A(X,A)$ is an isomorphism for separable finite algebras over fields. But the question remains unsolved in the general case.