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\$$.