Infinite group acts on a set such that an orbit of any length exists. 
Give an example of an infinite group $G$ which acts on a set $S$ such that for each $n\in \mathbb{N}$ there is an orbit of this length.

Has anyone got an idea?
I've been trying something rotations in $\mathbb{C}$ of powers of matrices, something with eigenvalues, ...
I haven't found any example yet..
 A: Take $S$ to be the disjoint union of $\mathbb{Z}/n$ on which acts $G=\mathbb{Z}$.
A: Take the subgroup of permutations of $\mathbb Z$ generated by a permutation that shifts each even number two to the right, and partitions the odd numbers into a cycle of length 1, a cycle of length 2, a cycle of length 3, ...
(The even numbers gives you an orbit of length $\infty$, which you may or may not need to provide).
A: Recall the classification of group actions: every group action is a disjoint union of its orbits, which are transitive group actions. A transitive group action is isomorphic to $G/G_x$ where $G_x$ is the stabilizer of any element, and every subgroup appears in this way.
Hence the possible sizes of orbits of actions of $G$ are precisely the possible indices of subgroups of $G$. So your question reduces to:

Find an example of a group $G$ with a subgroup of every possible finite index.

The easiest example of such a group is $G = \mathbb{Z}$ but many others are possible. 
A: You can definitely do this using the idea of complex rotations, as you were inclined to. The idea is that you want a linear transformation such that $e^{2i \pi/n}$ is an eigenvalue for every positive integer $n$, since the corresponding eigenvector has period $n$.
Let $S$ be the set of functions $f:\mathbb N\rightarrow \mathbb C$, where $\mathbb N$ is the set of positive integers. Let $\mathbb Z$ act on $S$ by saying that, $n\in\mathbb Z$ takes $f:\mathbb N\rightarrow\mathbb C$ to the function
$$n*f(k)=f(k)\cdot e^{2ni\pi/k}.$$
In particular, $1\in\mathbb Z$ it acts on the $k^{th}$ component of the function $f$ by multiplication by multiplication by $e^{2i\pi/k}$. Obviously, a function $f$ which is non-zero only at $k$ has period $k$ under this action.
