This is a problem from Rotman 5.37. I do not think I grasp the gist of the problem statement.
$R$ might not be a commutative ring here. A functor $F:\mathcal{A}\to\mathcal{B}$ is strong isomorphism if there is a functor $G:\mathcal{B}\to\mathcal{A}$ such that $GF=1_{\mathcal{A}},FG=1_{\mathcal{B}}$. Clearly $Hom_R(R,-):Mod_R\to Mod_R$ is naturally isomorphic to $1_{Mod_R}$ where $Mod_R$ is the right $R-$module category. Then $Hom_R(R,-)$ is not strong isomorphism. And it should be boring?
If $Hom_R(R,-)$ is strong isomorphism(note it is exact and the adjoint is also exact), then the adjoint of it should be $-\otimes R:Mod_R\to Mod_R$ as $R$ is $(R,R)$ bimodule. However, $R$ is flat as $(R,R)$ module which makes $-\otimes R= 1_{Mod_R}$ natural isomorphism. Apparently, that it is strong isomorphism from this.
Something is wrong with my understanding above.
I just concluded $Hom_R(R,-)=-\otimes R=1_{Mod_R}$. I do not think I grasped the gist of the problem.
When talking about "strong isomorphisms" (which are usually just called isomorphisms of categories), you need to be careful to distinguish between (naturally) isomorphic objects and equal objects.
In particular, for instance, if $F$ is a strong isomorphism, then $F$ must be surjective on objects, since $FG=1_\mathcal{B}$ is surjective on objects. This is not true of $F=Hom_R(R,-)$: not every module $M$ is equal to a module of the form $Hom_R(R,N)$. Here by "equal" I mean equal in the most literal possible set-theoretic sense (equal underlying set and equal module operations), not just isomorphic. For instance, by definition, any element of the underlying set of $Hom_R(R,N)$ is a function, so if you take a module $M$ whose underlying set has an element which is not a function, it cannot be equal to $Hom_R(R,N)$ for any module $N$.