Suppose $F:mathrm{Grp}tomathrm{Grp}$ is an equivalence. The object $mathbb{Z}inmathrm{Grp}$ is a minimal generator (it is a generator, and no proper quotient is also a generator), and this property must be preserved by equivalences. Since there is a unique minimal generator, we can fix an isomorphism $phi:mathbb Zto F(mathbb Z)$. Now $F$ must preserve arbitrary coproducts, so for all cardinals $kappa$, the isomorphism $phi$ induces an isomorphism $phi_kappa:L_kappato F(L_kappa)$, where $L_kappa$ is the free product of $kappa$ copies of $mathbb Z$. In particular, if $1$ is the trivial group, $phi_0:1to F(1)$ is an isomorphism.
Next pick a group $Ginmathrm{Grp}$, and consider a free presentation $L_1to L_0to Gto1$, that is, an exact sequence with the $L_i$ free. (For simplicity, we can take $L_0=L(G)$ the free group on the set underlaying to $G$, and $L_1$ to be the free group on the subsetunderlaying the kernel of the obvious map $L_1to G$; this eliminates choices) Since $F$ is an equivalence, we have another exact sequence $F(L_1)to F(L_0)to F(G)to F(1)$. Fixing bases for $L_1$ and $L_0$ we can use $phi$ to construct isomorphisms $L_ito F(L_i)$ for both $iin{0,1}$. Assuming we can prove the square commutes, one gets an isomorphism $phi_G:Gto F(G)$—this should not be hard, I guess.
The usual arguments prove then in that case the assignment $G mapsto phi_G$ is a natural isomorphism between the identity functor of $mathrm{Grp}$ and $F$.
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