Isomorphism of algebras vs isomorphism of categories of representations Let $A, B$ be finite-dimensional algebras over a field, so that their categories of modules $A\text{-mod}$ and $B\text{-mod}$ are finite abelian in the sense of EGNO Tensor categories.
Given an isomorphism $f:A \to B$ of algebras, the pullback $f^* : B\text{-mod} \to A\text{-mod}$ is an isomorphism of finite abelian categories. Furthermore, it commutes with the forgetful functors, in the sense that
\begin{align}
    \hom_A(A, f^*V) \cong \hom_B(B, V),
    \quad g \mapsto (b \mapsto b.g(1))
\end{align}
natural in $B$-modules $V$. Here, $A$ and $B$ are the regular modules. (actually, this isomorphism is
I was wondering if we can reverse this statement.
However, I don't see it, and I don't understand where isomorphism vs equivalence plays a role.
So far I did

*

*take $F: B\text{-mod} \to A\text{-mod}$ an equivalence of finite abelian categories

*such that there exists $\nu : \hom_A(A, F-) \overset{\sim}{\Rightarrow} \hom_B(B,-)$
We have the vector space isomorphism $FB \cong \hom_A(A,FB) \xrightarrow{\nu_B} \hom_B(B,B) \cong B$.
Note also that $F$ is an equivalence so $A \cong FGA$, where $G$ is a quasi-inverse of $F$.
In particular, then,
\begin{align}
    \hom_B(B,V) \cong \hom_A(A,FV) \cong \hom_A(FGA,FV) \cong \hom_B(GA,V)
\end{align}
whence by Yoneda $A \cong FB$ in $A\text{-mod}$.
In particular, $A \cong B$ as vector spaces.
I feel I'm overlooking something very obvious here. There at least should be some condition on $\nu$; and I guess $F$ should actually do nothing on morphisms?
Any hints would be appreciated.
 A: $\newcommand{\id}{\operatorname{id}}
\newcommand{\tensor}{\otimes}$
This follows  from Lemma 1.6 in Hopf Monads (Bruguieres, Virelizier).
Indeed, if $A$ is an algebra, then $A\otimes -$ is a monad on $\text{Vect}$, and the category of modules over the monad is precisely the category of modules over $A$.

EDIT:
I will now be more explicit.
Without loss we can actually assume that the forgetful functor sends a module to its underlying vector space (this functor is isomorphic to the one I wrote above).
Let's at first not restrict to equivalences/isomorphisms of categories.
So our initial set-up is the following:

*

*$F : B\text{-mod} \to A\text{-mod}$ a functor (+ appropriate adjectives) which commutes with the forgetful functors to $\text{Vect}$.

This already implies two things:

*

*$Ff = f$, i.e. $F$ doesn't do anything on intertwiners.

*$F(V,\rho) = (V,\hat{\rho})$ with $\hat{\rho} : A\otimes V\to V$, for a $B$-module $(V,\rho)$. [Note that we denote a module by a pair consisting of a vector space and an action]

Define for any vector space the linear map
\begin{align}
    f_{V} = \widehat{m_B \tensor \id_V} 
    \circ \id_A \tensor 1_B \tensor \id_V
    : A \tensor V \to B \tensor V \ ,
\end{align}
where $1_B$ is the unit of $B$.
Note that this is natural in $V$ and thus in fact determined by its value $f=f_k$ on the field, i.e. $f_V = f \tensor \id_V.$
Let $(V,\rho) \in B\text{-mod}$.
Consider the free $B$-module $(B \tensor V, m_B \tensor \id_V)$, where $m_B$ is the multiplication of $B$.
We can consider $\rho$ as a morphism in $B\text{-mod}$, from the free module above to $(V,\rho)$.
Then in particular $F\rho = \rho$ is an $A$-module map, which means
\begin{align}
    F\rho \circ \widehat{m_B \tensor \id_V}
    = \widehat{\rho} \circ \id_A \tensor F\rho \ .
\end{align}
Thus we find for the interplay between $\rho$ and $f_{V}$
\begin{align}
    F\rho \circ f_{V}
    &= F\rho \circ
    \widehat{m_B \tensor \id_V}  
    \circ \id_A \tensor 1_B \tensor \id_V
    \\
    &= \widehat{\rho} 
    \circ \id_A \tensor F\rho 
    \circ \id_A \tensor 1_B \tensor \id_V
    \\
    &=
    \widehat{\rho}
    \ .
\end{align}
That is, we have that $\widehat{\rho} = \rho \circ f_{V} = \rho \circ f \tensor \id_V$ as linear maps.
Finally, the claim that $F = f^*$ follows once one has proved that the following are equivalent for a linear map $f:A \to B$ between algebras:

*

*$f$ is an algebra map

*If $(V,\rho)$ is a $B$-module, then $(V, \rho \circ f \tensor \id_V)$ is an $A$-module.

I will leave this as a good exercise (if done correctly, the proof takes less than one A6-sized page)
