Let $H$ be a group, $\phi$ an automorphism of $H$ of order n and fix $h\_0 \in H$. I wonder, what the restrictions are, such that 
$$G:= \lt H,g \mid g^n=h\_0,\quad \forall h \in H: ghg^{-1}=\phi(h) \gt$$
defines a group which has $H$ as normal subgroup such that $G/H$ is cyclic of order n. 


There are two obvious restrictions: 


(1) $h\_0$ has to be in the center of $H$. For, $h\_0hh\_0^{-1} = g^nhg^{-n}=\phi^n(h) = h$, since $\phi$ has order n. 


(2) $\phi(h\_0) = h\_0$. For, $\phi(h\_0) = gh\_0g^{-1}=gg^ng^{-1} = g^n = h\_0$. 


But I can't figure out, if there some more restrictions. 


I tried to apply the classification of extensions with non-abelian kernel (Kenneth Brown, Cohomology of Groups, chapter IV, §6), but that requires to consider $H^3(Out(H),C)$ ($C$ the center of $H$) and I'm unable to do so, because I have no information about $Out(H)$ and $C$. 


Any help is appreciated.



Restrictions for presenting groups with cyclic quotient 
Your conditions are sufficient. Indeed, consider the $H$-by-cyclic group $G\_0=\langle H, g\mid ghg^{-1}=\phi(h), h\in H\rangle$. By (1) and (2), both $g^n$ and $h\_0$ are central in that group. Hence the element $u=h\_0^{-1}g^n$ is central. Now factor out the central subgroup $\langle u\rangle$: $G=G\_0/\langle u\rangle$. That group $G$ is what you need. 


