Contravariant functor taking every finite group to an isomorphic group

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 $F$.

@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:

1. $F(G) \cong G$ (not neccesarily naturally) for all finite groups $G$, and
2. $F^2 : \mathsf{FinGrp} \to \mathsf{FinGrp}$ 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 $A_5 \to \mathbb{Z}/2\mathbb{Z}$.