The strategy has two parts: (1) verify that the $n$ rotations and $n$ reflections are isometries of the polygon, then (2) prove that every isometry of the polygon is one of these $2n$ maps, using a rigidity argument based on distance preservation. **Step 1: The rotations and reflections are isometries.** Place the regular $n$-gon with vertices at $V = \{e^{2\pi ik/n} : 0 \leq k < n\} \subset \mathbb{C}$. Define the rotation $r(z) = e^{2\pi i/n} z$ and the reflection $t(z) = \bar{z}$. For the rotation: \begin{align*} |r(z) - r(w)| = |e^{2\pi i/n}(z - w)| = |e^{2\pi i/n}||z - w| = |z - w|. \end{align*} For the reflection: \begin{align*} |t(z) - t(w)|^2 = |\bar{z} - \bar{w}|^2 = (\bar{z} - \bar{w})(z - w) = |z - w|^2. \end{align*} Both preserve distances, so they are isometries. Since $r$ permutes $V$ cyclically and $t$ permutes $V$ (sending $e^{2\pi ik/n}$ to $e^{-2\pi ik/n}$), the $2n$ maps $\{e, r, r^2, \ldots, r^{n-1}, t, rt, r^2t, \ldots, r^{n-1}t\}$ are all isometries of the polygon. **Step 2: Every isometry has this form.** [claim:Rigidity of Polygon Isometries] Let $f$ be an isometry of the polygon. Then $f \in \{e, r, \ldots, r^{n-1}, t, rt, \ldots, r^{n-1}t\}$. [/claim] [proof] Since $f$ permutes $V$, we have $f(1) = e^{2\pi ik/n}$ for some $k$. Set $g = r^{-k} \circ f$, so $g(1) = 1$. Since $g$ is a composition of isometries, it is an isometry. The vertex $e^{2\pi i/n}$ is at a fixed distance from $1$, and $g$ must send it to a vertex at the same distance from $g(1) = 1$. The only such vertices are $e^{2\pi i/n}$ and $e^{-2\pi i/n}$. **Case 1:** $g(e^{2\pi i/n}) = e^{2\pi i/n}$. Then $g$ fixes two adjacent vertices. By induction around the polygon (each subsequent vertex is determined by its distances to the two already-fixed neighbours), $g$ fixes all vertices. So $g = \mathrm{id}$, giving $f = r^k$. **Case 2:** $g(e^{2\pi i/n}) = e^{-2\pi i/n}$. Then $t \circ g$ fixes both $1$ and $e^{2\pi i/n}$. By the same inductive rigidity argument, $t \circ g = \mathrm{id}$, so $g = t$ (since $t^2 = \mathrm{id}$), giving $f = r^k \circ t$. [/proof] **Step 3: Conclusion.** The $2n$ maps are distinct (they induce distinct permutations of $V$), and every isometry is among them, so the symmetry group has exactly $2n$ elements. Closure under composition follows from the relations $r^n = t^2 = e$ and $trt = r^{-1}$, which can be verified directly. This [group](/page/Group) is the dihedral group $D_{2n}$.