Consider a finite-dimensional algebra $A$ over a field $k$. We have the dualization functor $*: Mod-A \to A-Mod$ given by
$$M^*=Hom_k(M,k).$$
This gives us a left $A$-module $A^*$. In general, it does not have to be isomorphic to $A$ as a left module.
But we may consider $A^*$ as a right module over the opposite algebra $A^{op}$. What if $A^{op}$ is isomorphic to $A$? Given a fixed isomorphism $A\cong A^{op}$, is there a way to give a canonical isomorphism $A^*\cong A$ as left $A$-modules (=right $A^{op}$-modules)? This seems like such a triviality, but I have been lost in all the isomorphisms and reversing multiplication orders for a few hours now.
I'm not sure if this statement is true. The semi-canonical choices of $k$-linear morphisms $A \to A^*$ that I can come up with fail to be left $A$-module homomorphisms.
Nevertheless I will describe here my attempt. Let us choose a basis $e_1, \dots, e_n$ for $A$ and let us simply spell out the condition for a $k$-linear morphism $\psi: A \to A^*$ to be a morphism of left $A$-modules. As a basis for $A^*$ we choose $\{\hat{e}_i\}_{i=1}^n$, where \begin{aligned} \hat{e_i}(e_j) := \begin{cases} 1 & i = j; \\ 0 & \text{otherwise.}\end{cases} \end{aligned} Let $\phi: A \to A^{\rm op}$ be the given involution of $A$, i.e. $\phi(1) = 1$ and $\phi(ab) = \phi(b)\phi(a)$. As you indicated, we then make $A^*$ into a left $A$-module by \begin{aligned} (a \cdot f)(b) := f(b\phi(a)). \end{aligned} Let's spell out what this is in coordinate. The multiplication $\mu: A \otimes A \to A$ can be written out in coordinates as \begin{aligned} e_i e_j = \mu(e_i \otimes e_j) = \sum_{k} \mu_{ijk} e_k. \end{aligned} Similarly the involution $\phi: A \to A^{\rm op}$ can be written in coordinates as \begin{aligned} \phi(e_i) = \sum_l \phi_{il} e_l. \end{aligned} The left $A$-action on $A^*$ thus becomes \begin{aligned} (e_i \cdot \hat{e}_j)(e_k) = \hat{e}_j(e_k\phi(e_i)) = \sum_m \phi_{im} \hat{e}_j(e_k \cdot e_m) = \sum_m \phi_{im} \mu_{kmj}, \end{aligned} or in other words \begin{aligned} e_i \cdot \hat{e}_j = \sum_m \sum_k \phi_{im} \mu_{kmj} \hat{e}_k. \end{aligned} Now let us assume there is a $k$-linear isomorphism $\psi: A \to A^*$ given by $\psi(e_i) = \sum_j \psi_{ij} \hat{e}_j$ such that $\psi$ is an isomorphism of left $A$-modules. Then $\psi$ should satisfy the equation \begin{aligned} \psi(ab) = a \cdot \psi(b), \end{aligned} or in other words \begin{aligned} \psi(ab)(c) = \psi(b)(c\phi(a)). \end{aligned} Writing both sides out in coordinates, one obtains \begin{aligned} \psi(e_ie_j)(e_k) = \psi(\sum_{l}\mu_{ijl}e_l)(e_k) = \sum_l \mu_{ijl} \psi_{lk} \end{aligned} and \begin{aligned} \psi(e_j)(e_k\phi(e_i)) = \sum_m \phi_{im} \psi(e_j)(e_k e_m) = \sum_m \sum_n \phi_{im} \mu_{kmn} \psi(e_j)(e_n) = \sum_m \sum_n \phi_{im} \mu_{kmn} \psi_{jn}. \end{aligned} So all in all we need to find $\psi$ satisfying \begin{aligned} \sum_l \mu_{ijl} \psi_{lk} = \sum_m \sum_n \phi_{im} \mu_{kmn} \psi_{jn}. \end{aligned} We see that taking $\psi_{ij} = \delta_{ij}$ could work whenever we have the identity \begin{aligned} \mu_{ijk} = \sum_m \phi_{im} \mu_{kmj}, \end{aligned} which seems like a very strange condition to me. Another attempt could be taking $\psi_{ij} = \phi_{ij}$, which probably also won't work in general since the condition $\phi(ab) = \phi(b)\phi(a)$ translates into the condition that \begin{aligned} \sum_l \mu_{ijl} \phi_{lk} = \sum_m \sum_n \phi_{jn} \phi_{im} \mu_{nmk}; \end{aligned} the above condition would thus for example be satisfied if $\mu_{nmk} = \mu_{kmn}$ were to hold, which again seems like a strange condition.
I wouldn't be surprised at this point when someone comes up with an explicit counterexample.