To explain what seems to be there you need to describe it more exactly. It might help to draw an edge from $i$ to $j$ starting a little after $i$ because the edges do have a direction.

For the graphs $n/(n+1)$ there is an edge from $i$ to $n+1-i$ (so we could in this case consider them undirected.) This gives a reflective symmetry of order $2$ but it can't be a purely rotational symmetry unless $n+1$ is even (in which case it is). It is true that the edges are all horizontal (if $0$ is at the top) so in the case that $n+1$ is odd, overlaying the diagram with itself rotated a half turn puts each rotated edge in a position half way between two parallel former edges of similar length.

So you need to be more specific about what you mean (in your answer below) by $a+b$-fold symmetry when the number of points $an+b$ is not a multiple of $a+b.$

I will look at the graphs $n/(an+b)$ in the case that $a$ and $b$ are relatively prime and $an+b$ is a multiple of $a+b.$ Then $n \bmod (a+b) \equiv 1$ so

$$n/(an+b)={\Big(}p(a+b)+1{\Big)}/{\Big(}(ap+1)(a+b){\Big)}.$$ Here the $(a+b)$ fold rotation would take $i$ to $i+ap+1.$

Note that $$ n(ap+1)={\Big(}p(a+b)+1{\Big)}(ap+1) \equiv ap+1 \bmod {\Big(}(ap+1)(a+b){\Big)} $$

So the edge from $i$ to $ni$ does indeed rotate to an edge from $i+ap+1$ to $ni+n(ap+1) \equiv ni+ap+1.$

The case that $\gcd(a,b)=g \gt 1$ is similar.

**LATER**

My advice wasn't to draw more graphs, it was to be more specific about the exact nature of the seeming approximate similarity. Ideally expressing it symbolically and using arithmetic/number theory to explain it.

Here is a case that should give sufficient detail to allow a more general analysis. I'll start with diagrams, describe a feature, and then explain it arithmetically. The actual history was in the other direction but it doesn't make as good a story.

I will look at the exact or "almost" $5$-fold symmetry For $n/(2n+3).$ It is better to do $n/(n+2)$ or $n/(2n+1)$ first but this case illustrates an additional feature.

Here, first is the case $16/35$ which is typical for $(5q+1)/(10q+5)$

The small outward segments at $0,7,14,21,28$ correspond to the "edges $(i,i)$ in those five cases. And we already understand the exact symmetry from a rotation of $\frac{7}{35}=\frac15.$

A good illustration of what you see is $17/37.$ The middle picture superimposes in red a rotation of $\frac15$ which looks approximately symmetric and on the right a rotation of $2/5$ which is not so symmetric.

But then there is $14/31.$ Here the rotation by $\frac15$ is not as close to a symmetry as the rotation by $2/5.$

But what does that mean?The complete edge list is

$[0, 0], [1, 14], [2, 28], [3, 11], [4, 25], [5, 8], [6, 22], [7, 5], [8, 19], [9, 2]$

$ [10, 16], [11, 30], [12, 13], [13, 27], [14, 10], [15, 24], [16, 7], [17, 21], [18, 4], [19, 18], [20, 1]$

$[21,15], [22, 29], [23, 12], [24, 26], [25, 9], [26, 23], [27, 6], [28, 20], [29, 3], [30, 17]$

There is an angle made by the edges $[5,8]$ and $[8,19]$ . Moving by $\frac25$ of a complete rotation would increase everything by $\frac{62}5=12.4$ and move that angle to $[17.4,20.4]$ and $[20.4,31.4]=[20.4,0.4]$ The closest actual edges are $[17,21]$ and $[20,1].$ These cross internally though close to the boundary.

This is not an isolated example. One can see that the edges add: $[6,22]+[7,5]=[13,27]. $ It is easy to see why. There is only one fixed point $[0,0]$ and the next closest $[12,13]$ and $[19,18].$ So an angle $[i,j],[j,p]$ rotates to $[i+12.4,j+12.4],[j+12.4,p+12.4]$ which is fairly close to actual (crossing) edges $[i+12,j+13],[j+12,p+13].$ It could be argued that it is slightly better to advance by $12.5$ which is $\frac{12.5}{31}=0.4032258$ of a rotation rather that $\frac25=0.4.$

So here is my conclusion for the case $n/m=n/(an+b)$ which I will limit, for now, to the case that $\gcd(n,m)=1.$ Then there is a unique $0 \lt k \lt m$ so that the edge $[k,nk]=[k,k+1] \bmod m.$ The "best" rotation is by $\frac{k+0.5}{m}$ (Rotating that much in the opposite direction is just as good due to the reflective symmetry.) Evidently this is pretty close one of $\frac{c}{a+b}$ and $c$ depends on the congruence class of $n \bmod {a+b}.$ The next step would be to specify how that works and prove it.

In the case $n/(2n+3):$

When $n=5q+2,$ such as $17/37$ above, one has $n/(2n+3)=(5q+2)/(10q+7)$ and for $k=2q+1,$we have $nk=10q^2+9q+2 \equiv 2q+2 \bmod m.$

When $n=5q+4,$ such as $14/31$ above, one has $n/(2n+3)=(5q+4)/(10q+11)$ and for $k=4q+4,$ $nk=20q^2+36q+16 \equiv 14q+16 \equiv 4q+5 \bmod m.$

The other two congruence classes, $5q$ and $5q+3,$ are similar.