Do there exist groups $G,H$ and $I$ together with group homomorphisms $h : G \rightarrow H$ and $i : G \rightarrow I$ such $\mathrm{img}(h)$ and $\mathrm{img}(i)$ are normal subgroups of $H$ and $I$ respectively, and $H/\mathrm{img}(h) \cong I/\mathrm{img}(i)$, but $H$ and $I$ are not isomorphic?
(Sorry, there was meant to be an assumption that $h$ and $i$ are injective in the original question. In any event, all concerns have been addressed by the answers below.)
On
I think the simplest example is to take $G = \Bbb Z_2, H = \Bbb Z_2\times \Bbb Z_2, I = \Bbb Z_2$, let $h(a) = (a, 0)$ and $i(a) = 0$.
On
Just take $G=H$ non-trivial, with $h$ the identity map, and take $I$ to be the trivial group.
Then $\operatorname{im}(h)=H$, $\operatorname{im}(i)=I$, and clearly both $H/H$ and $I/I$ are trivial.
(For more fancy examples, where $I$, $I/\operatorname{im}(i)$ and $H/\operatorname{im}(h)$ are non-trivial, fix your favourite group $K$ and notice that the above example "embeds" into $G':=G\times K$, $H':=H\times K$ and $I':=K$. Here, $I'$, $I'/\operatorname{im}(i')$ and $H'/\operatorname{im}(h')$ are all isomorphic to $K$.)
Yes. Take $G=\mathbb{Z}_3$, $H=\mathbb{Z}_3\times\mathbb{Z}_2$, $I=S_3$, $h(g)=(g,0)$, and $i(g)$ such that $i(1)=(1\ \ 2\ \ 3)$. Obviously, $H$ and $I$ are not isomorphic. But $H/\operatorname{im}(h)\simeq I/\operatorname{im}(i)\simeq\mathbb{Z}_2$.