Let $G$ be a finite simple group, and $F$ a profinite group (I'm really interested in the case where $F$ is free of finite rank, in particular rank 2).

In Ribes-Zalesskii, they define the $G$-rank of $F$ to be the integer $n_G$ such that if $K_G$ is the intersection of all open normal subgroups of $F$ whose quotient is isomorphic to $G$, then $F/K_G \cong G^{n_G}$.

At first, I thought this is the same as the largest integer $m_G$ such that $G^{m_G}$ is a quotient of $F$, but I've realized that this may not be so.

I believe (by a Goursat's Lemma argument) that $m_G$ is the cardinality of the set $Surj(F,G)/Aut(G)$.

Certainly, by taking kernels, every equivalence class in $Surj(F,G)/Aut(G)$ determines an open normal $U\le F$ with $F/U = G$, and hence $m_G\ge n_G$, but it's unclear to me if you can have two distinct classes in $Surj(F,G)/Aut(G)$ which have the same kernels.

If this is possible, has there been any work regarding when $n_G = m_G$, and if they're different, then how large the gap between $n_G$ and $m_G$ can be?