John asked in [comment #19](

> **Puzzle.** What is this obvious thing?

I think that the functor \\(\mathbf{op}(F)\\) behaves the same way as the functor \\(F\\) on both objects and morphisms:

- On objects, \\(\mathbf{op}(F) : \mathrm{ob}(\mathcal{C}^{\mathrm{op}}) \to \mathrm{ob}(\mathcal{D}^{\mathrm{op}})\\).
But objects in the opposite category are the same objects as in the original category, that is, \\(\mathrm{ob}(\mathcal{C}^{\mathrm{op}}) = \mathrm{ob}(\mathcal{C})\\);
so the functor \\(\mathbf{op}(F) : \mathrm{ob}(\mathcal{C}) \to \mathrm{ob}(\mathcal{D})\\) can map objects the same way as \\(F\\) does.
- On morphisms, \\(\mathbf{op}(F) : \mathcal{C}^{\mathrm{op}}(C', C) \to \mathcal{D}^{\mathrm{op}}(\mathbf{op}(F)(C'), \mathbf{op}(F)(C))\\) for any \\(C, C' \in \mathrm{ob}(\mathcal{C})\\).
The opposite category reverses the morphisms, that is, \\(\mathcal{C}^{\mathrm{op}}(C', C) = \mathcal{C}(C, C')\\);
so the functor \\(\mathbf{op}(F) : \mathcal{C}(C, C') \to \mathcal{D}(FC, FC')\\) can map morphisms the same way as \\(F\\) does.

So, I would be tempted to say that the \\(\mathbf{op}\\) functor sends each functor \\(F\\) to itself, that is, \\(\mathbf{op}(F) = F\\).