\\(F^\mathrm{op}\\) should do the same thing as \\(F\\) on objects (i.e., \\(\forall c\in\mathcal{C}\,\,F(c)=F^\mathrm{op}(c)\\)). On morphisms, I imagine it should take \\(f^\mathrm{op}\\) to \\((F(f))^\mathrm{op}\\). This works trivially on identities. \\(F^\mathrm{op}(f^\mathrm{op}\circ g^\mathrm{op})=F^\mathrm{op}((g\circ f)^\mathrm{op})\\) \\(=(F(g\circ f))^\mathrm{op}=(F(g)\circ F(f))^\mathrm{op}\\) \\(=(F(f))^\mathrm{op}\circ(F(g))^\mathrm{op}\\) \\(=F^\mathrm{op}(f)\circ F^\mathrm{op}(g)\\), so \\(F^\mathrm{op}\\) defined in this way is a functo from \\(\mathcal{C}^\mathrm{op}\\) to \\(\mathcal{D}^\mathrm{op}\\).