Picking out a subset of elements from a finite product of cyclic groups Let $C_n$ be the cyclic group of order $n$, and let $G = \prod_{i=1}^n C_n = \underbrace{C_n \times C_n \times \ldots \times C_n}_{n \text{ times}}$. 
For $g = (g_1,g_2,\ldots, g_n) \in G$, call $g$ a star iff $g_i \neq g_j$ for all $i,j$ where $i \neq j$. (There are many more elements than stars; the ratio of elements to stars is ${n^n}/{n!}$)
I am working with $G$ as part of a larger problem and am interested in finding some principled way (ideally using group-theoretic concepts and properties, ideally repeated multiplication by a group element from some other group $H$) to visit all stars and no non-stars. The starting point is immaterial, it could be $e \in G$ or a non star or any star, though ideally $e$ or a star. If there are no group-theoretic ways of doing this, I might define a function $f$ such that 
\begin{equation}f(g) =  \left\{
\begin{array}{cl}
g+(1,2,...,n) & \text{if } g= e,\\
g+ (1,1,...,1) & \text{if } g \text{ a star and } f^2(g) \text{ not previously visited},\\
\vdots & \text{if } etc.
\end{array} \right.\end{equation}
I know I cannot find a $g\in G$ which satisfies this condition. At most the orbit of some $g$ will contain $\phi(n)$ stars, where $\phi$ is Euler's totient function. If no such $h \in H$ exists, I think I must use $f$. This seems possible, though $f$ will become quickly awkward as $n$ becomes large. 
I am not a group theorist. Is there something obvious I don't know? Are there any results which inform my problem in any way?  
(BTW, I am not married to groups, I could set the problem in a ring or field if that makes things easier. For example I was thinking of defining $n$  continuous functions $x_i(t)$ such that for integral values of $t$ the $t^{th}$ star (for some ordering of $G$) is such that $x_i(t) = g_i$.) 

I resolved my problem in the following way: 
Let $M$ be some ordered collection of all permutations of $C_n$. So the first element of $M$ might be $M_1 = (1,2,\ldots, n)$, the second $M_2 = (2,1,\ldots n)$ and so on (here $n = e$). Let $f_i(t)$ be the sequence formed by taking the $i^{th}$ element from the $t^{th}$ element of $M$, where $t$ ranges from $1$ to $n$. 
Now we can form the function $f$ mentioned above by collecting the $n$ functions $f_i(t)$. Because $t$ is discrete we can analyze these functions using the discrete Fourier transform. 
