I was trying to answer this question in a way that it was sufficient to take $H$ such that $\phi_1$ could be chosen to be injective. At first I thought of taking $H=G_1$ and $\phi_1=id_{G_1}$, but I don't know how to define $\phi_2$ in a way such that $\phi_1\circ i_1=\phi_2\circ i_2$.
I don't know if this is even possible, so can I always choose such $H$?
Taking $\phi_1=id_{G_1}$ does not work in general.
If we take $i_1=id_A$, then the condition on the homomorphisms reduces to $id_{A_1}=\phi_2 \circ i_2$. But this means that $i_2$ has to admit a left inverse.
If $i_2:\mathbb Z \to \mathbb Z$, $1\mapsto 2$, then there is no left inverse.
Remark/abstract nonsense: In the example I used that $\mathbb Z $ is not divisible. Divisble abelian groups are exactly the injective abelian groups and being injective is a necessary condition on the group $G_2$ if we want $\phi_1$ to be the identity.