Given two categories, $C$ and $D$, a covariant functor is usually defined as a regular functor $C \to D$, whereas a contravariant functor is usually defined as a regular functor $C^{op} \to D$.
Question: Could we also define a contravariant functor (equivalently) as a functor $C \to D^{op}$?
Or is there an isomorphic natural transformation between functors $C \to D^{op}$ and $C^{op} \to D$?
I am trying to think of what the difference would be (if any).
The answer is yes, both formulations are equivalent. Let's see the correspondence.
Let's see that giving a contravariant functor $F:C^{op}\to D$ is equivalent to giving another functor $G:C\to D^{op}$.
Assume we know $F$. At the level of objects it's clear what to do: for any $a\in Ob(C)=Ob(C^{op})$ we define $G(a):=F(a)\in Ob(D^{op})=Ob(D)$.
Now, given any arrow $g\in\hom_C(a,b)=\hom_{C^{op}}(b,a)$, note that $\hom_D(F(b),F(a))=\hom_{D^{op}}(F(a),F(b))$, so we may regard $F(g)$ as an element of $\hom_{D^{op}}(F(a),F(b))$ via this identification, therefore we can define $G(g)=F(g)$. Clearly $G$ is a functor since $F$ is.
The other equivalence is pretty similar, so I'll let you do it.