Contravariant functor taking every finite group to an isomorphic group 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.
 A: This was a hard one - but a really nice question so worth it.
Let $a$ denote a rotation of $\mathbb{R}^2$ about the origin through $24^\circ$ (so $a$ has order $15$) and let $b$ denote a reflection across a line through the origin.  Then we have dihedral groups generated as follows:
\begin{eqnarray*}
D_{30}&=&\langle a,b\rangle,\\
D_{6}&=&\langle a^5,b\rangle,\\
D_{10}&=&\langle a^3,b\rangle,
\end{eqnarray*}
where by construction we have inclusions $\iota_1\colon D_6\hookrightarrow D_{30}$ and $\iota_2\colon D_{10}\hookrightarrow D_{30}$.
Let $f\colon D_{30}\to D_6 \times D_{10}$ denote the group homomorphism sending:
\begin{eqnarray*}
a&\mapsto& (a^{10},a^6),\\
b&\mapsto&(b,b).
\end{eqnarray*}
Let $p_1,p_2$ denote the obvious projections from $D_{6}\times D_{10}$ to $D_6$ and $D_{10}$ respectively.  Then consider the following diagram:
\begin{align*}\begin{array}{ccccc}
&&D_6&&\\ &\stackrel {\iota_1}\swarrow && \stackrel {\quad p_1}\nwarrow\qquad&\\
&D_{30}&\stackrel f\longrightarrow &D_6\times D_{10}\\
&\stackrel {\qquad\iota_2}\nwarrow && \stackrel{\quad p_2\qquad}\swarrow\qquad&\\
&&D_{10}&&
\end{array}
\end{align*}
We have:
\begin{eqnarray*}
p_1f\iota_1&=&1_{D_6},\\
p_2f\iota_2&=&1_{D_{10}}.\\
\end{eqnarray*}
Now suppose that $F$ is a contravariant functor from the category of finite groups to itself, such that $FG$ is isomorphic to $G$ for all finite groups $G$. We have:
\begin{eqnarray*}
(F\iota_1)(Ff)(Fp_1)&=&1_{D_6},\\
(F\iota_2)(Ff)(Fp_2)&=&1_{D_{10}}.\\
\end{eqnarray*}
Thus the image of $Ff$ must contain groups isomorphic to $D_6$ and $D_{10}$.  Thus $|{\rm im} (Ff)|=30$ and $|{\rm ker} (Ff)|=2$.  In particular, $D_6\times D_{10}$ must contain a normal subgroup of order $2$.  In other words $D_6\times D_{10}$ contains a central element of order $2$.  However it contains no such element so we have a contradiction to the existence of $F$.
A: @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 $A_5 \to \mathbb{Z}/2\mathbb{Z}$.
