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If $N(H)$ is normalizer of $H$, then the following example should work. $G:=\mathbb{Z}\times\mathbb{F}_2$ and $H$ is a subgroup $(e,H_0)$, where $H_0$ is of finite index but not normal in $\mathbb{F}_2$. Then index of $N(H)=\mathbb{Z}\times N_{\mathbb{F}_2} (H_0)$ is finite and not equal to $1$, since $N_{\mathbb{F}_2}(H_0)\neq \mathbb{F}_2$, because $H_0$ is not normal in $\mathbb{F}_2$.

If $N(H)$ is normal closure of $H$ in $G$, then the following example should work. Let $\mathbb{F}_2$ be the free group on two generators $a$ and $b$. Define a homomorphism $\phi:\mathbb{F}_2\rightarrow Aut(\mathbb{Z})=\mathbb{Z}_2$ by $\phi(a)=0,\phi(b)=1$. Then the question is valid for the semidirect product $G=\mathbb{F}_2\ltimes_{\phi} \mathbb{Z}$. The Schreier graph of $G/H$ has polynomial growth and the normal closure of $H$ is $\mathbb{F}_2\ltimes_{\phi} 2\mathbb{Z}_2$.