TITLE: Contravariant functor taking every finite group to an isomorphic group
QUESTION [7 upvotes]: Every finite abelian group $A$ is isomorphic to its dual group $A^*:=\operatorname{Hom}(A,\mathbb{C}^\times)$. The isomorphism of $A$ with $A^*$ is non-canonical, and one way to make this precise is to say that the functor $A\mapsto A^*$ is contravariant, so this functor cannot be naturally isomorphic to the (covariant) identity functor. I wonder if there is an analogous construction that works for all finite groups. Specifically:

Does there exist a contravariant functor $F$ from the category of finite groups to itself, such that $F(G)$ is isomorphic to $G$ for every group $G$?

The imprecise question I have in mind is: for an arbitrary finite $G$, can one construct a group $G'$ that is non-canonically isomorphic to $G$? The existence of a contravariant functor $F$ as above would be one precise way to answer the imprecise question.

REPLY [4 votes]: @ikf's answer is excellent, but I want to point out that while the isomorphism $A^* \cong A$ isn't canonical, there is a canonical natural isomorphism $A \cong (A^*)^*$, given by $a \mapsto (f \mapsto f(a))$. It's much easier to show that there is no functor on all finite groups satisfying this additional constraint. To be precise:
Claim There is no functor $F : \mathsf{FinGrp}^{\text{op}} \to \mathsf{FinGrp}$ such that:

$F(G) \cong G$ (not neccesarily naturally) for all finite groups $G$, and
$F^2 : \mathsf{FinGrp} \to \mathsf{FinGrp}$ is naturally isomorphic
to $\operatorname{id}_{\mathsf{FinGrp}}$.

Proof. Since $F^2$ is naturally isomorphic to $\operatorname{id}_{\mathsf{FinGrp}}$, it is faithful. This implies that $F$ is faithful. This is impossible because there are $16$ homomorphisms $\mathbb{Z}/2\mathbb{Z} \to A_5$, but only one homomorphism $A_5 \to \mathbb{Z}/2\mathbb{Z}$.