Suppose you have two $k$ algebras $A, B$ (say also finitely generated if this helps) and a functor $F: A-mod \to B-mod $ such that $| F(M) |= |M|$. Here $|U|$ denotes the underlying $k$ vector space.


Can you find sufficient conditions so that $F$ is the restriction functor relative to $f:B \to A$?


I suspect that a "nice" condition to start from is monoidality. Indeed, in the case in which the rings are the group rings of finite groups, by Tannaka reconstruction you could reconstruct the morphism.


My motivation: I had to find an $\mathbb{R}$-algebra injective morphism from $\mathbb{C} \to \mathbb{R}[u]/(u^2+1) ^m$. It's really easy with Jordan form theory to give a complex structure on a module for the right hand ring, but constructing the actual morphism it's not easy at all, and some non trivial ad-hoc argument is needed to make " reconstruction ".



Functors between module categories that comes from restriction
Let's assume that we are using right modules and that the isomorphism of underlying vector space functors is a natural isomorphism because we will just work up to isomorphism of functors. Note that $F$ is exact. To come from restriction $F$ should preserve all limits and colimits. Maybe this follows by functorially preserving underlying spaces but I'm not sure. If it doesn't you probably want to require that. It follows from Eilenberg-Watts that there is an $B$-$A$-bimodule $M$ such that $F\cong \hom\_A(M,-)$. Moreover, since $A$ represents the underling vector space functor, we must have that $M\cong A$ as a right $A$-module. Since $End\_A(A)\cong A$, it follows the left $B$-module structure comes from a homomorphism $F\colon B\to A$. So basically $F\cong \hom\_A(A,-)$ via this bimodule structure. But then evaluating at $1$ gives an isomorphism of $F$ with the functor taking $N$ to $N$ with $B$-action $mb = mf(b)$. So $F$ is isomorphic to restriction along $f$.


