The quotient of the affine space $\mathbb{A}^n$ by the symmetric group $Sym\_n$ is again an affine space of the same dimension, and invariants are given by elementary symmetric polynomials. 


What about a $n$-cycle? More precisely, what are the generators of the invariants of the action of $\mathbb{Z}/n\mathbb{Z}$ on $\mathbb{A}^n$ given by $(x\_1,\dots,x\_n)\mapsto (x\_2,x\_3,\dots,x\_n,x\_1)$ ?


It is probably classical, but I was not able to find it.



Quotient of affine space by cyclic permutation
Rita's answer is perfectly correct. However, Jérémy posed his question over a (not necessarily algebrically closed) characteristic $0$ field
$k=\mathbb{Q}$, so here is another approach. Consider the subring $S = k[s\_1,\dots,s\_n]$ generated by the elementary symmetric polynomials, i.e., the invariant ring under the entire symmetric group, not just the cyclic group. The entire polynomial ring $R = k[x\_1,\dots,x\_n]$ is an $S$-algebra that is, in fact, a 
free $S$-module of rank $n!$. The fact that this is a free $S$-module follows automatically since $S$ is regular and $R$ is Cohen-Macaulay; however, there are also perfectly explicit choices of $S$-basis, e.g.,
the monomials $\underline{x}^{\underline{e}} =x\_1^{e\_1}x\_2^{e\_2}\cdots x\_n^{e\_n}$ indexed by all exponents $(e\_1,\dots,e\_n)\in \mathbb{Z}\_{\geq 0}^n$ with $e\_i < i$ for every $i$. The $k$-algebra homomorphism $$\phi:R\to R, \ \ \phi(x\_1,\dots,x\_{n-1},x\_n) = (x\_2,\dots,x\_n,x\_1),$$ is an $S$-algebra homomorphism. The ring of $\phi$-invariants is the kernel of the $S$-module homomorphism $\text{Id}\_R-\phi:R\to R$, which is the same as the image of the "averaging homomorphism" / Maschke homomorphism / Reynolds operator $$E=\frac{1}{n} ( \text{Id} + \phi + \phi^2 + \dots + \phi^{n-1} ).$$
In particular, the subring of $\phi$-invariants is a direct summand of $R$ as an $S$-module. Of course one set of generators is the set of images $E(\underline{x}^{\underline{e}})$ as above; 
you can compute a smaller set of $S$-generators for the kernel using a term order on $R$ as an $S$-module and using Gröbner basis methods. Any collection of $S$-generators give $k$-algebra generators once you add in the $k$-algebra generators of $S$, i.e., the elementary symmetric polynomials.


Regarding the question when the ring of invariants is a polynomial ring, or even a smooth $k$-algebra, this fails for all $n\geq 3$. As explained in the comments, this can be checked after base change to $\overline{k}$, where it becomes an immediate corollary of the Chevalley-Shephard-Todd theorem (or just direct computation). However, as you can see, the invariant ring is Cohen-Macaulay; it is a free $S$-module (special case of the Hochster-Roberts Theorem). 


$\textbf{Edit}$. It seems quite plausible that, as a free $S$-module, the ring of invariants has a basis consisting of $E(\underline{x}^{\underline{e}})$ for those $(n-1)!$ exponents as above satisfying the additional condition that $e\_n=0$. 


