Geometric Intuition for Dihedral Group Automorphisms

I noticed the other day that the automorphism group of the dihedral group $D_{2n}$ (of order $2n$) is $\operatorname{Aff}(\mathbb Z/n\mathbb Z)$, the group of affine transformations of the $\mathbb Z$-module $\mathbb Z/n\mathbb Z$. By "affine transformation" I mean an invertible function $\mathbb Z/n\mathbb Z\to \mathbb Z/n\mathbb Z$ of the form $x\mapsto ax + b$ (so $a\in(\mathbb Z/n\mathbb Z)^\times$ and $b\in\mathbb Z/n\mathbb Z$); the group multiplication in $\operatorname{Aff}(\mathbb Z/n\mathbb Z)$ is function composition. Both $D_{2n}$ and $\operatorname{Aff}(\mathbb Z/n\mathbb Z)$ have a natural geometric interpretation: $D_{2n}$ represents the symmetries of the $n$-gon and $\operatorname{Aff}(\mathbb Z/n\mathbb Z)$ represents a nice class of transformations of the "space" $\mathbb Z/n\mathbb Z$. The geometric interpretation of $\operatorname{Aff}(\mathbb Z/n\mathbb Z)$ is admittedly more abstract than the first, but I don't think it should be discounted. At any rate, I understand the algebraic reasons why $\operatorname{Aut}(D_{2n}) \cong \operatorname{Aff}(\mathbb Z/n\mathbb Z)$, but it seems like there might also be geometric reasons.

Is there an intuitive geometric explanation for why we should expect the automorphism group of a dihedral group to be an affine group? Does anyone know of an insightful way to visualize automorphisms of the dihedral group as affine transformations?

Picture $D_{2n}$ as the symmetries of the unit circle $S^1 \subset \mathbb{C}$ marked by the $n^{th}$ roots of unity $\omega_b = \text{exp}(2 \pi i b/n)$, $b \in \mathbb{Z} / n \mathbb{Z}$.

The short answer to your question is that $\text{Aff}(\mathbb{Z} / n \mathbb{Z})$ acts by $n$-fold self-covering maps of $S^1$ that permutes the lines of reflective symmetry of the elements of $D_{2n}$, and that this action is identical to the action of $\text{Aut}(D_{2n})$ on the lines of reflective symmetry. To me, what this is saying is that automorphisms of $D_{2n}$ are realized by "self-covering maps" of $D_{2n}$.

The $n$ rotational symmetries are the maps $\rho_b(z) = z \omega_b$. The $n$ reflective symmetries are the reflections $r_c$ across the lines of the form $L_{c}$, where $L_c$ passes through the origin and through the $(2n)^{\text{th}}$ root of unity $\text{exp}(2 \pi i c/2n)$, where $c \in \mathbb{Z} / 2n \mathbb{Z}$. We have a one-to-one correspondence $r_c \leftrightarrow L_c$ between reflection elements reflection lines: each reflection determines its line, each line determines its reflection. The group $\text{Aut}(D_{2n})$ permutes the set of reflections $r_c$, and this is faithful action meaning that an automorphism is completely determined by its reflection permutation (because the reflections generate $D_{2n}$). Therefore $\text{Aut}(D_{2n})$ acts faithfully by permutations on the set of reflection lines $L_c$.

Enter $\text{Aff}(\mathbb{Z}/n\mathbb{Z})$. The claim is that there is a faithful $\text{Aff}(\mathbb{Z}/n\mathbb{Z})$ action on the set of reflection lines $L_c$ which is exactly the same as the action of $\text{Aut}(D_{2n})$. Given an element $(a,b) \in \text{Aff}(\mathbb{Z}/n\mathbb{Z})$, you can apply it to the circle by an $a$-fold covering map according to the formula $$(a,b) \cdot z = z^a \cdot \omega_b $$ When applying this to $z = exp(2 \pi i c / 2n)$ the result is $z' = exp(2 \pi i (ac + 2b)/2n)$ and so you get $$(a,b) \cdot L_c = L_{c'} \,\,\text{where}\,\, c' = ac+2b $$ and from this formula it easily follows that the action of $\text{Aff}(\mathbb{Z}/n\mathbb{Z})$ on the lines $L_c$ is a bijection.

There's still some computation to do in order to verify that the actions of $\text{Aff}(\mathbb{Z}/n\mathbb{Z})$ and of $Aut(D_n)$ on the reflection lines $L_c$ are the same, but at least I've given the underlying geometric idea.

Here's a way to visualize these so-called "covering maps". Fix $n$ and fix $(a,b) \in \text{Aff}(\mathbb{Z}/n\mathbb{Z})$ (try, for example, $n=5$, $a=2$, and $b=$anything in $\mathbb{Z}/5\mathbb{Z}$). Consider the following relabelling of the points $\omega_0,\omega_1,\ldots,\omega_{n-1}$: let $\xi_0 = \omega_b$, and inductively let $\xi_i = \xi_{i-1} \cdot \omega_a$. Starting with $\xi_0$, connect a straight line to $\xi_1$, then $\xi_2$, ..., then $\xi_{n-1}$, and then back to $\xi_n=\xi_0$ (in the example you get a pentagram, with vertices labelled $0,1,2,3,4$). The automorphism that corresponds to $(a,b)$ has the following effects: it takes the reflection fixing $\omega_i$ to the reflection fixing $x_i$; and it takes the reflection across the bisector of the segment $\overline{\omega_i \omega_{i+1}}$ to the reflection across the bisector of the segment $\overline{x_i x_{i+1}}$.