Let $G$ and $H$ be arbitrary groups, and then let $f, j \colon G \to H$ be homomorphisms. How can I then prove that $\ker f = \ker j$ if and only if $\exists \rho \colon H \to H$, an automorphism of H, such that $Image(j) = \rho (Image(f))$? I already know how to prove that the same kernel implies isomorphic images, but I don't know how to prove the converse.
For a little background on my experience, I am a novice self-studying group theory from Artin's Algebra, although this question is mine and mine alone (meaning that I did not find it or anything related to it in my textbook).
This is not true. Consider $G = H = \mathbb{Z}$, $f(x) = x$, and $j(x) = 2x$. Then $\ker f = \ker j = 0$, and $im(f) = \mathbb{Z}$, while $im(j) = 2 \mathbb{Z}$. It’s easy to see no automorphism of $\mathbb{Z}$ induces an isomorphism between $im(f)$ and $im(j)$.
Note that this isn’t even true if we switch to talking about vector spaces - it’s nothing unique to groups.