Initially, I wanted to ask this question as a puzzle.

Consider a regular $m$-gon. Let $0$ be the lower corner and count the corners clockwise.

Let $n_m$ be the multiplication-by-$n$-graph of $m$ for $n < m$ by drawing an arrow from every corner $a$ to the corner $a\cdot n\operatorname{mod}m$.

_{These are the eight relevant multiplication-by-$n$-graphs of $10$ as (hopefully) intelligible to every school kid:}For fixed $n$, in the limit $m \rightarrow \infty$ the graph $n_m$ will

somehowdisplay $n-1$ cusps._{E.g. $10_{517}$:}What is the smallest graph $n_m$ that

undeniablydisplays $n-1$ cusps?

You may think hard about this (or not), but the (probably right) answer can be found easily by simulation: Just draw the graphs

and you will find that the graph with $m = n^2-1$

yields this smallest graph.

Be promised: It's the same for arbitrary $n$ instead of $10$:

Even for smaller $n$ than $10$:

My question is:

Whyis this so?

I'm looking for strong algebraic and/or arithmetic arguments. I have no idea.