I was learning the Krull-Schmidt theory and came across this concept and just can't understand what's it all about.
A group endomorphism $f\colon G\to G$ is called normal iff $f(aba^{-1})=af(b)a^{-1}$ for all $a,b\in G$. It's true that $H$ is a normal subgroup of $G$ implies $f(H)$ is a normal subgroup of $G$, given that $f$ is a normal endomorphism on $G$.
Is the converse true? E.g. Is it true that "an endomorphism $f$ on group $G$ images every normal subgroup of $G$ to a normal subgroup" implies "$f$ is a normal endomorphism"?
If it's not true, some other way to understand this definition would be appreciated(what does it have to do with normality?).
Hint If $f$ and $H$ are normal, then for all $g \in G$ you have $$ gf(H)g^{-1}=\{ gf(h)g^{-1} : h \in H \} = \{ f(ghg^{-1}) : h \in H \} \subseteq f(H) $$ Therefore, $f(H)$ is normal subgroup.
The converse is not true. The simplest counterexample is $f: A_5 \to A_5$ defined by $$f(x)=gxg^{-1}$$ for some $g\in S_5$ which we will pick later. Since $A_5$ is simple, $f$ trivially maps normal subgroups into normal subgroups.
Now, if $a\in A_5$ then setting $b=a$ you have $$f(aba^{-1})=af(b)a^{-1} \Leftrightarrow \\ gag^{-1}=agag^{-1}a^{-1} \Leftrightarrow \\ (gag^{-1})a=a(gag^{-1}) $$
Now, all you have to do is find some $g\in S_5,a \in A_5$ such that $gag^{-1}$ does not comute with $a$. This is easy, pick $a$ a 5-cycle, and pick some $g$ such that $gag^{-1}$ is not a power of $a$.