On weakly symmetric algebras with Nakayama involution Let $A$ be a weakly symmetric $k$-algebra, that is, $A$ is a finite-dimensional self-injective $k$-algebra such that $\mathrm{top}  P \cong \mathrm{soc} P$ for any indecomposable projective $A$-module $P$.
Let $\nu$ be the Nakayama functor, that is, $\nu = D(A) \otimes_A \_: mod(A) \to mod(A)$, where $D(A) = Hom_k(A,k)$ denotes the dual of $A$.
Assume that $\nu$ is an involution, that is, assume that there is an isomorphism $\nu^2 \cong Id_{mod(A)}$ of functors.
Is it true that for any left $A$-module $M$ there is an isomorphism $\nu(M) \cong M$?
This isomorphism is not required to be functorial in $M$.
Such an isomorphism exists for any projective and any simple $A$-module.
Moreover, there is an isomorphism of functors $\nu \cong Id_{mod(A)}$ if and only if $A$ is symmetric. 
The answer is positive in the case of the local anti-commutative algebra $A = k \langle x,y \rangle /(x^2,y^2, xy + yx)$. 
Any comments would be highly appreciated.
EDIT: Without the assumption $\nu^2 \cong Id_{mod(A)}$, there are examples with non-$\nu$-invariant modules (in "Frobenius Algebras I" by Skowronski and Yamagata, Chapter IV, Section 10, Example 7).
 A: Let $A$ be any algebra where $\nu_A^2\cong\text{Id}$ but $\nu_A\not\cong\text{Id}$, such as the example $A=k\langle x,y\mid x^2=y^2=xy+yx=0\rangle$ that you give in the question.
So, as $A$-bimodules, $A\cong D(A)\otimes_A D(A)$ but $A\not\cong D(A)$.
Let $TA$ be the trivial extension algebra $TA=A\oplus D(A)$ with multiplication $(a,\varphi)(a'\varphi')=(aa',a\varphi'+\varphi a')$. All I'll need to know about $TA$ is that it is symmetric and has $A$ as an algebra quotient.
Let $B=A\otimes_k(TA)^{op}$, so a left $B$-module is just an $A$-$TA$-bimodule.
$$D(B)\otimes_B D(B)\cong \left(D(A)\otimes_A D(A)\right)\otimes_k \left(TA\otimes_{TA}TA\right)\cong B$$
as a $B$-bimodule, since $TA$ is symmetric. So $\nu_B^2\cong\text{Id}$.
Let $M=A$, considered as a $B$-module, or $A$-$TA$-bimodule, by left multiplication of $A$ and right multiplication of $TA$ via its natural quotient $A$.
Then 
$$\nu_B(M)\cong D(A)\otimes_A M\otimes_{TA}D(TA)\cong D(A)\otimes_AA\cong D(A)$$
as an $A$-$TA$-bimodule. 
But this is not isomorphic to $M=A$ as a bimodule, since $DA\not\cong A$ as an $A$-bimodule. 
So the algebra $B$, with module $M=A$, is a counterexample.
Probably there's a much more elementary example along the lines of $A=k\langle x,y\mid x^2=y^2=xy+yx=0\rangle$ where you could do an explicit calculation, but I tend to get horribly confused by any calculation that involves keeping track of minus signs.
