Give an Example of a Bijective Function $\mathbb{Z} \rightarrow \mathbb{Z}$ with precisely $r$ orbits This problem is a continuation of what was discussed here. 
As the title states, I wish to find a bijective function $\mathbb{Z} \rightarrow \mathbb{Z}$ with precisely $r$ orbits. I observe from the definition of an orbit $O_a$ that, essentially, each of its members depend on the preimage $a$. Hence, if the domain is infinitely large, in this case $\mathbb{Z}$ is $\aleph_0$ large, then there will be just as many orbits (again $f$ is bijective, so this seems to be a safe conclusion regardless of how $f$ is defined). For this reason, a bijective function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ with infinitely many orbits is simply the identity mapping.
Assuming my observations are correct, I cannot see how a bijective function with $\mathbb{Z}$ as its domain can have finitely many orbits, unless for an arbitrary $r \in \mathbb{N}$, we take a finite nonempty subset $V \subseteq \mathbb{Z}$ such that $|V| = r$ as the domain and codomain of the function $j: V \rightarrow V$ defined by $j(v) = v$ for all $v \in V$. It would then follow that the number of orbits depend on the cardinality of V, namely the arbitrary natural number $r$.
Any insight or assistance would be greatly appreciated! Thanks in advance!

Map any integer $z$ to $z+r$. The congruence classes modulo $r$ are the orbits.
Remark: In the OP, the comment was made that the only bijective function with infinitely many orbits is the identity function. There are many more. For example, we can use the function $f$ that for all $z\in \mathbb{Z}$ interchanges $2z$ and $2z+1$.
