Equivalence of $R$-linear additive categories are $R$-linear? Let $R$ be a commutative ring. Let $C,D$ be $R$-linear categories (https://stacks.math.columbia.edu/tag/09MJ ) such that they are also additive categories. Let $F: C \to D$ be an equivalence of categories.
Then, is $F$ an $R$-linear functor? i.e., for every $X,Y\in  Ob(C)$, is it true that
$$F: \operatorname{Hom}_C(X,Y) \to \operatorname{Hom}_D(F(X), F(Y))$$
is an $R$-module homomorphism?
My thoughts: Since $C,D$ are additive categories, so $F$ is an additive functor i.e. $F: \operatorname{Hom}_C(X,Y) \to \operatorname{Hom}_D(F(X), F(Y))$
is an abelian group homomorphism. So I think we only have to check whether $F: \operatorname{Hom}_C(X,Y) \to \operatorname{Hom}_D(F(X), F(Y))$
preserves multiplication by elements of $R$ or not.
 A: No.  For a really simple way of getting examples, note that if $\alpha:R\to R$ is any endomorphism and $C$ is an $R$-linear category, we can make a different $R$-linear category $D$ with the same underlying category as $C$ but with the action of $R$ modified via $\alpha$ (i.e., $r\cdot_D f=\alpha(r)\cdot_C f$).  Then the identity functor is an equivalence $C\to D$, but it is will rarely be $R$-linear.  In particular, if $\alpha$ is not the identity and the action of $R$ on $C$ is faithful (e.g., if $C$ is the category of $R$-modules), the identity functor is not $R$-linear.
Or for another really simple type of example (if you don't require the categories to be additive, as the original question did not), note that an $R$-linear category with one object is just an $R$-algebra.  Then you are asking whether if $A$ and $B$ are $R$-algebras and $f:A\to B$ is an isomorphism of multiplicative monoids, then $f$ is $R$-linear.  There are very easy counterexamples to this (for instance, when $R=A=B=\mathbb{Z}$).
