Are algebras with invertible linear duals always Frobenius? Let $A$ be a finite dimensional algebra over a ground field $k$. The linear dual $A^* = Hom_k(A,k)$ is naturally an $A$-$A$ bimodule. I am interested in those algebras such that $A^*$ is an invertible $A$-$A$ bimodule. That is, there is another $A$-$A$ bimodule $L$ and $A$-$A$ bimodule isomorphisms $L \otimes_A A^* \cong A \cong A^* \otimes_A L$.
One class of algebras which has this property are the Frobenious algebras. One of the classical definitions of a Frobenius algebra is that it is an algebra with an isomorphism of right $A$-modules ${A^*}_A \cong A_A$. If this is an isomorphism of bimodules then this is a symmetric Frobenius algebra. More generally we have ${}_A{A^*}_A \cong {}_A{}^\sigma A_A$, where the right-hand side is simply $A$ as a bimodule but where the left action is twisted by the Nakayama isomorphism $\sigma$. In particular since the Nakayama isomorphism is an isomorphism, $A^*$ is an invertible bimodule.
Question: If $A$ is an algebra such that $A^*$ is an invertible bimodule, does $A$ admit the structure of a Frobenius algebra?
Upon reviewing some old notes to myself, apparently at one time I believed that the answer to the above question is yes. However I don't remember the reasoning and didn't record a reference. Further, I am suspicious of my old self because in general there are certainly invertible bimodules which do not come from twisting the left action of the trivial bimodule. I would be happy to understand a counterexample or to find out that my old self was right.
One motivation for studying these algebras is that they arise naturally in extended topological field theory. There is a certain variant of 2D framed tqfts (the "non-compact" variant) and these algebras are in bijection with those tqfts with values in the Morita 2-category. So I would also be interested in anything further that could be said about these algebras, even with further assumptions like $k$ being characteristic zero.
 A: For a finite dimensional algebra $A$, $A^{\ast}$ being an invertible
bimodule is equivalent to $A$ being self-injective (which is the same
as quasi-Frobenius for finite dimensional algebras).
One implication has already been covered in comments. If $A^{\ast}$ is
invertible, then $-\otimes_{A}A^{\ast}$ is a self-equivalence of the
right module category, and so sends projectives to projectives. So
$A^{\ast}$ is projective.
For the other implication, assume $A$ is self-injective. Then
$-\otimes_{A}A^{\ast}$ is left adjoint to
$\operatorname{Hom}_{A}(A^{\ast},-)$, and it is easy to check that
the unit
$$A\to \operatorname{Hom}_{A}(A^{\ast},A\otimes_{A}A^{\ast}),$$
which is given by $a\mapsto[\varphi\mapsto a\otimes\varphi]$ for
$a\in A$, $\varphi\in A^{\ast}$, is an isomorphism.
But $\operatorname{Hom}_{A}(A^{\ast},-)$ is exact and therefore
isomorphic to $-\otimes_{A}L$, where
$L=\operatorname{Hom}_{A}(A^{\ast},A)$, by the Eilenberg-Watts
theorem. So $A^{\ast}\otimes_{A}L\cong A$ as $A$-bimodules.
The same argument with left modules shows that $A^{\ast}$ has a left
inverse, and so $A^{\ast}$ is invertible.
For a typical example of a self-injective algebra that is not
Frobenius, start with a Frobenius algebra $A$ with an indecomposable
projective right module $P$ such that $P\otimes_{A}A^{\ast}\not\cong
P$, and take a Morita equivalent algebra $B$ that is the endomorphism
algebra of a progenerator that contains $P$ and $P\otimes_{A}A^{\ast}$
as direct summands with different multiplicities.
The simplest example is where $A$ is the path algebra of a quiver with
two vertices $v_{1}$ and $v_{2}$, with an arrow $a$ from $v_{1}$ to
$v_{2}$ and an arrow $b$ from $v_{2}$ to $v_{1}$, modulo the relations
$ab=0=ba$. Let $e_{i}$ be the idempotent corresponding to vertex
$v_{i}$, and $P_{i}=e_{i}A$ the corresponding indecomposable
projective right module.
Then $B=\operatorname{End}_{A}(P_{1}^{2}\oplus P_{2})$ is
self-injective (since it's Morita equivalent to $A$) but not
Frobenius. The indecomposable projective corresponding to $P_{1}$
under the Morita equivalence occurs with multiplicity two as a summand
of $B$, but with multiplicity one as a summand of $B^{\ast}$.
