Does $F/N_{1}\cong F/N_{2}$ implies $N_{1}\cong N_{2}$ and vice versa? It would be great for me to see an example of finite group $F$ and two epimorphisms $\varphi_{i}:F\twoheadrightarrow G$ from $F$ onto a group $G$, with $N_{1}\ncong N_{2}$, where $N_{i}=ker\varphi_{i}$, $i=1,2$ (e.g., finite group which contains two non-isomorphic subgroups each of index $2$).
However, I am interested more in the case $F=F(a,b)$ the free group on the set $\{a,b\}$. That is, the case we have two isomorphic presentations $\langle a,b|R_{1}\rangle\cong\langle a,b|R_{2}\rangle$ but with non-isomorphic normal closures $N_{i}=N_{F}(R_{i})$, $i=1,2$. Note that if $G_{1}$ (or $G_{2}$) is finite, then the ranks of $N_{1}$ and $N_{2}$ are equal, so they are isomorphic (since they are free by Nieslen-Schreier).
What about the converse: if $N_{1}\cong N_{2}$ are two isomorphic normal subgroups of the free group $F=F(a,b)$, can we deduce that $F/N_{1}\cong F/N_{2}$?
Thanks.

For example, let $G=D_8$ of order $8$, generated by $a$ of order $4$ and $b$ of order $2$. Then $\langle a\rangle$ and $\langle a^2,b\rangle$ are normal subgroups of index $2$, non-isomorphic. (One is cyclic of order $4$, one is Klein four.)
For free groups, every subgroup of a free group is free, so there are three options for $N_1$. It has finite rank, infinite rank, or is trivial. It has finite rank if and only if it has finite index. Thus if $|F/N_1|$ is finite then $F/N_1\cong F/N_2$ implies $N_2$ has finite index.
The rank of $N_i$ is given by a formula in terms of index and number of generators of the free group ($n(d-1)+1$, where $d$ is the number of generators and $n$ the index), and so $N_1\cong N_2$ in this case. If $N_1\cong N_2$ then all you know is that $F/N_1$ and $F/N_2$ have the same order.
If $N_1$ is trivial then $F/N_1\cong F$. As Derek Holt says, this means that $N_2=1$ as $F$ is not isomorphic to a proper quotient of itself.
The final case is when $N_1$ is not finitely generated, or equivalently, $N_1$ has infinite index (and is non-trivial). Then all such $N_i$ are isomorphic (as $F$ is countable).
Thus if $F/N_1\cong F/N_2$ then $N_1\cong N_2$, and if $N_1\cong N_2$ then $F/N_1$ and $F/N_2$ have the same order.
