If an endofunctor F on some category $\mathfrak{C}$ is (left) adjoint to the identity functor.... then does it necesarilly have to also be the identity functor... Since $\mathfrak{C}(F(X),Y)\cong \mathfrak{C}(X,1_{\mathfrak{C}}(Y))=\mathfrak{C}(X,Y)$. ... So $F(X)\cong 1$.
(so, unless I'm wrong .. wouldn't that provide another proof for $-\otimes_R R:_RMod \rightarrow _RMod$ being the identity endofunctor $1_{_RMod}$? If we accept the fact that $Hom_R(R,-)\cong 1_{_RMod}$), or am I making the mistake in assuming that this $Hom_R(-,?)$ is exact?
Any invertible functor $F$ has both a left and a right adjoint, namely its inverse. Left and right adjoints are unique up to (unique) natural isomorphism, so the correct statement is that if $F$ is adjoint (on either side) to the identity, then it is naturally isomorphic to the identity.