The option 4. is false because $(\mathbb{Q},+)$ has no elements of finite order other than $0$, whereas every element of $S_3$ has finite order.
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.
Best Answer
$\{0\}$ is the only finite subgroup of $(\Bbb Q,+)$, and hence there is no nontrivial homomorphism from any finite group $G$ to $(\Bbb Q,+)$.