I am not sure how I can draw a commutative diagram, so I will do my best to describe it verbally.
So, suppose $f_1,f_2:G\to K$ be (group, field, ring)homomorphisms. I want to claim the following:
- There exists an isomorphism $g:G\to G$ such that $f_1\circ g = f_2$, and
- Such isomorphism $g$ is unique.
Question 1: Is it even a true statement?
Question 2: If not, what other conditions to I need to add to make it so?
Thanks bunch in advance.
This is an extremely untrue statement in general: if the images $f_1$ and $f_2$ are different, it is clear that there cannot be any such $g$, unique or otherwise.
The only situation I can think of where something similar to your statement is true is with normal field extensions. From Lang's Algebra, p.237:
The condition he calls "NOR 1" means that when $K/k$ is a normal field extension with $K\subseteq k^a$, for any map $f:K\to k^a$ (of $k$-algebras - it's not sufficient to consider them only as fields), the image of $f$ will always be $K$, so that any such map can be expressed as $f=i\circ \sigma$ where $\sigma\in\operatorname{Aut}(K/k)$ and $i:K\to k^a$ is the inclusion of $K$ as a subset.
Thus if $K/k$ is a normal field extension and $f_1,f_2:K\to k^a$ are two maps of $k$-algebras, there is some $\sigma\in\operatorname{Aut}(K/k)$ such that $f_2=f_1\circ \sigma$, and this $\sigma$ would be unique because $\operatorname{Aut}(K/k)$ is a group.