This is a nice exercise...Hints
1) $\,\operatorname{Aut}(N)\,$ is abelian
2) Every inner automorphism of $\,G\,$ is, when restricted to $\,N\,$ , is an element of $\,\operatorname{Aut}(N)\,$
3) $\,\forall x,y\in G\,\,,\,[x,y]^{-1}=[y,x]\,$
Try now to do something with this and, if after thinking it over for a while you're still stuck, write back below as a comment.
Added on request: As noted, $\,\operatorname{Aut}(N)\,$ is abelian and if $\,\phi_g\,$ denotes the inner automorphism determined by $\,g\,$ , then $\,\forall\,g\in G\,\,\,,\,\,\text{then}\;\; \left.\phi_g\right|_N\,\in\operatorname{Aut}(N)$ . We show now that any basic commutator $\,[x,y]\in H\,$ centralizes any element $\,n\in N\,$ :
$$[x,y]n[x,y]^{-1}=[x,y]n[y,x]=x^{-1}y^{-1}xyny^{-1}x^{-1}yx=\left(\phi_{x^{-1}}\phi_{y^{-1}}\phi_x\phi_y\right)(n)=$$
$$\stackrel{\text{Aut}(N)\,\,\text{is abelian!}}=\left(\phi_{x^{-1}}\phi_x\phi_{y^{-1}}\phi_y\right)(n)=Id_N\circ Id_N(n)=n$$
and since the above is true for any generator of $\,H=G'=[G,G]\,$ then it is true for the whole group.
Second solution: Perhaps easier: for any subgroup $\,K\leq G\,$ , the map $$f:N_G(K)\to\operatorname{Aut}(K)\,\,,\,\,f(k):=\phi_k=\,\text{conjugation by}\,\,k$$
is a group homomorphism (with $\,\phi_k(x):=kxk^{-1}\,$), whose kernel is precisely $\,C_G(K)\,$ , and from here
$$N_G(K)/C_G(K)\cong T\leq\operatorname{Aut}(K)$$
In our case, we have $\,N\triangleleft G\Longleftrightarrow N_G(N)=G\,$ , so that we get $\,G/C_G(N)\cong T\leq\operatorname{Aut}(N)\,$ .
But $\,\operatorname{Aut}(N)\,$ is abelian, so that
$$G/C_G(N)\,\,\,\text{is abelian}\,\,\Longleftrightarrow G'\leq C_G(N)\;\;\;\;\;\;\square$$
The fact is that in general the question is false, but if one add the hypothesis,
that the group $G$ is nilpotent (e.g. a $p$-group for some prime $p$), then the result become true:
Indeed, we have the chain $\gamma_{3}(G)\subset \gamma_{2}(G)\subset G$,
where $\gamma_{2}(G)= G'$ and $\gamma_{3}(G)=[\gamma_{2}(G), G]$ (the commutator). Now from the last relation we have $\frac{\gamma_{2}(G)}{\gamma_{3}(G)}=Z( \frac{G}{\gamma_{3}(G)} ) $ (the center);
so the quotient $\frac{\frac{G}{\gamma_{3}(G)}}{Z( \frac{G}{\gamma_{3}(G)} ) }=\frac{G}{\gamma_{2}(G)}$ and this is cyclic, and as someone have remarked above, this implies that $\frac{G}{\gamma_{3}(G)}$ is abelian and then $G'=\gamma_{2}(G)\subset \gamma_{3}(G)$ and finally $\gamma_{2}(G)= \gamma_{3}(G)$. Since $\gamma_{3}(G)=[\gamma_{2}(G),G]=[\gamma_{3}(G),G]=\gamma_{4}(G)$ and so on... But in last we know that $G$ is nilpotent and so there is a $c \in \mathbb{N}^{+}$ such that $\gamma_{c}(G)=1$ and so $G'=1$ that means $G$ is abelian.
For the general case one can see to the alternating group $A_{4}$:
indeed one can prove that $|A_{4}|=12$, $|A'_{4}|=4$ so that the quotient $\frac{A_{4}}{A'_{4}}$ has 3 element and it is cyclic; but $A_{4}$ is not abelian.
Best Answer
The commutator is defined as $[a,b]=aba^{-1}b^{-1}$. Since the group is cyclic, it's Abelian. So, we have that $\forall a,b: ab=ba$. This gives that $aba^{-1}b^{-1}=e$ always. So, the commutator subgroup is $e$.