Suppose $F:\mathrm{Grp}\to\mathrm{Grp}$ is an equivalence. The object $\mathbb{Z}\in\mathrm{Grp}$ 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 $\phi:\mathbb Z\to F(\mathbb Z)$. Now $F$ must preserve arbitrary coproducts, so for all cardinals $\kappa$, the isomorphism $\phi$ induces an isomorphism $\phi_\kappa:L_\kappa\to F(L_\kappa)$, where $L_\kappa$ is the free product of $\kappa$ copies of $\mathbb Z$. In particular, if $1$ is the trivial group, $\phi_0:1\to F(1)$ is an isomorphism.

Next pick a group $G\in\mathrm{Grp}$, and consider a free presentation $L_1\to L_0\to G\to1$, that is, an exact sequence with the $L\_i$ free. (For simplicity, we can take $L\_0=L(G)$ the free group on the underlying set of $G$, and $L_1$ to be the free group on the underlying subset the kernel of the obvious map $L_1\to G$; 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 $L_i\to F(L_i)$ for both $i\in\{0,1\}$. *Assuming* we can prove the square commutes, one gets an isomorphism $\phi_G:G\to F(G)$—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 $F$.