This is Exercise 4.9 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search and Approach0, it is new to MSE.
The Details:
From page 140 of Roman's book,
Definition: Let $G$ be a group. A homomorphism $\sigma:G\to H$ is normality-preserving if $$N\unlhd G\implies \sigma N\unlhd H.$$
On the same page,
Definition: An endomorphism $\sigma:G\to G$ of a group $G$ is normal if $\sigma$ commutes with all inner automorphisms $\gamma_g$ of $G$, that is, for any $a\in G$, $$\sigma(a^g)=(\sigma a)^g$$ for all $g\in G$.
NB: The latter is the only definition of a homomorphism being normal that I could find in the book so far.
The Question:
Are all normality-preserving homomorphisms normal? Hint: Use the fact that $S_3$ is simple.${}^{\dagger}$
NB: I think the question should be, "are all normality-preserving endomorphisms normal".
Thoughts:
I suspect that the answer is "no" but I can't prove it yet. The group $S_3$ might be a counterexample.
I think I need to use the fact that $S_3$ is the smallest of the nonabelian groups.
The identity map ${\rm id}_{S_3}$ is normality-preserving trivially and normal.
The map
$$\begin{align} \varphi :S_3 & \to S_3,\\ \tau &\mapsto e \end{align}$$
is an endomorphism, and, since $\{e\}\unlhd \{e\}$, $\varphi$ is normality-preserving. It is also clearly normal.
Cheating a bit, the endomorphism monoid of $S_3$ has ten elements${}^{\dagger\dagger}$. They're given, I suppose, by their actions on the generators, so consider the presentation
$$S_3\cong\langle a,b\mid a^3, b^2, bab=a^{-1}\rangle\tag{1}$$
of $S_3$. I'm not sure what to do next (besides list all ways $a$ and $b$ can be mapped to elements in $S_3$; but surely there's a more sophisticated way than that, right?). Presentations aren't covered in the book so far too.
Please help :)
$\dagger$: But $S_3$ is not simple . . .
$\dagger\dagger$: See the Groupprops page "Endomorphism structure of symmetric group:S3". (I couldn't link to it because the address has a close-bracket.)
The question as stated has an easy answer: “no”, because “normal” is a property of endomorphisms, and normality preserving morphisms need not be endomorphisms.
So presumably, the actual question should be: “Is every normality preserving endomorphism a normal morphism?”
As suggested by Moishe Kohan, let $\sigma$ be a normal endomorphism. Given $g\in G$, we know that $\sigma(a^g) = \sigma(a)^g$; i.e., $$\sigma(g)^{-1}\sigma(a)\sigma(g) = g^{-1}\sigma(a)g,$$ and therefore, that $$(\sigma(g)g^{-1})^{-1}\sigma(a)(\sigma(g)g^{-1}) = \sigma(a).$$ Therefore, for every $g\in G$, $\sigma(g)g^{-1}\in C_G(\sigma(G))$; it centralizes the image of $\sigma$.
You can construct an example with an automorphism of $S_3$ (the nontrivial endomorphism that are not automorphisms have image of order $2$, so they are never normality-preserving). For instance, consider conjugation by $(12)$. This is normality preserving, but not normal: because $\sigma(13) = (12)(13)(12) = (123)$, and $\sigma(13)(13)^{-1} = (123)(13) = (23)\notin Z(S_3)$, so it is not normal.
P.S. My first counterexample was with $Q_8$, using the map $i\mapsto j\mapsto k\mapsto i$; I picked $Q_8$ because every subgroup is normal, so every endomorphism is normality preserving.