Let $G$ a group and let a group $\Gamma=G\times G$ with $\forall (a,b),(c,d)\in\Gamma, (a,b)(c,d):=(ac,bd)$. Let a map $\varphi:G\to\Gamma$ by $\varphi(g)=(g,g)$. Prove that $Z(G)=\left \langle e \right \rangle$ if and only if $N_\Gamma (Im(\varphi))=Im(\varphi)$.
Define $Q:=\operatorname{Im}(\varphi)$. I've shown that $Q=\{(g,g):g\in G\}$, $\varphi$ is a homomorphism and that $G$ is an abelian group if and only if $Q$ is a normal subgroup of $\Gamma$.
Suppose $Z(G)=\langle e\rangle$ and let $(a,b)\in N_\Gamma(\operatorname{Im}(\phi))$. Then for all $g\in G$ we have
$$\operatorname{Im}(\phi)\ni (a,b)(g,g)(a,b)^{-1}=(aga^{-1},bgb^{-1}),$$
which (by your calculation of $\operatorname{Im}(\phi)$) means that for some $g'\in G$ we have
$$(aga^{-1},bgb^{-1})=(g',g')\implies aga^{-1}=bgb^{-1}.$$
Rearrange this equation to see that $a^{-1}b$ commutes with $g$, and since this holds for every $g\in G$ we have $a^{-1}b\in Z(G)=\langle e\rangle$, which implies $a=b$ so $(a,b)\in \operatorname{Im}(\phi)$.