A divisor $d$ of a natural integer $N$ defines a permutation
of $\{0,\ldots,N-1\}$ by considering
$$x\longmapsto \pi_{d\vert N}(x)=\left\lfloor \frac{x}{d}\right\rfloor+\frac{N}{d}
\left( x\pmod d\right)$$
for $x \pmod d$ in $\{0,\ldots,d-1\}$.
One can show that the group $A(N)$ generated by all permutations $\pi_{d\vert N}$
associated to divisors of $N$ is abelian. (The group
$A(N)$ is in fact generated by $\pi_{p\vert N}$ for $p$ running through
all prime divisors of $N$.)
Example: $A(p^k)$ is cyclic of order $k$ if $p$ is prime.
It seems that $A(N)$ is fairly small:
I have no example of $N\geq 3$ such that $A(N)$ has more than $N-2$ elements.
(It is however not cyclic
in general.)
What does the group $A(N)$ measure?
Best Answer
All the $\tau$'s fix $0$ so we can look at how they behave on the set $\{1,2,\dots,N-1\}$. We have $$\tau_{d|N}(kd+r)=k+\frac{rN}{d}=\left[\frac{N}{d}(kd+r)\right]_{\pmod{N-1}}$$ where $[a]_{\pmod{N-1}}$ is the unique integer in $\{1,2,\dots, N-1\}$ congruent to $a\pmod{N-1}$. So $\tau_{d|N}$ is the same as multiplication by $\frac{N}{d}$ modulo $N-1$.
Therefore we see that the group $A(N)$ is the subgroup of $(\mathbb Z/(N-1)\mathbb Z)^{\times}$ generated by the divisors of $N$, therefore $$|A(N)|\le \varphi(N-1)\le N-2$$ as predicted by the OP.