Relating functors to relational functors with the parametricity translation
$(F, F^R)$ is not necessarily a relational functor. Define $F : \text{Set} \to \text{Set}$ to be the identity functor on sets and functions, but let ${F_!}^R$ send all relations to the trivial relation. Then ${F_{\text{map}}^R}$ trivially holds (along with ${F_{\text{id}}^R}$ and ${F_{\text{comp}}^R}$ which always hold for any $F$, because of the proof-irrelevance of equalities), but $F^R$ does not preserve identity relations.
Going back to the linked question: if a functor is defined internally in a parametric type theory, we don't really care about whether identity extension is derivable: we always get it as part of the &quot;free theorem&quot; package, so every internal $\text{Type} \to \text{Type}$ functor is also a relational functor externally.