Rigid motions taking a regular $n$-gon back to itself carry vertices to vertices

I have been reading Keith Conrad's expository paper Dihedral groups I and I have two questions about Theorem $$2.2$$, which deals with the size of $$D_n$$. In the first part of the proof you can read

An element $$g$$ of $$D_n$$ is a rigid motion taking the $$n$$-gon back to itself, and it must carry vertices to vertices (how are vertices unlike other points in terms of their distance relationships with all points on the polygon?) and $$g$$ must preserve adjacency of vertices.

I think I understand why $$g$$ must preserve adjacency of vertices: if $$A$$ and $$B$$ are vertices, then $$g(A)$$ and $$g(B)$$ are vertices, and the distance of $$g(A)$$ from $$g(B)$$ equals the distance of $$A$$ from $$B$$. Therefore, $$g(A)$$ and $$g(B)$$ are adjacent. Is this reasoning correct?

I think (but I'm not sure) that proving that $$g$$ preserves adjacency of vertices requires the fact that $$g$$ carries vertices to vertices. Why is that true? In particular, what is the answer to the author's question? (How are vertices unlike other points in terms of their distance relationships with all points on the polygon?) Does this question bear any relation to Lemma 2.1 in Conrad's notes?

EDIT: I'm still confused. I don't completely understand @zipirovich's answer: how can I rigorously show that the maximum $$d(P)$$ exists and that the value of $$d(P)$$ is largest when $$P$$ is a vertex, and smaller for any other point in the polygon?

Maybe my comment is a possible answer, but I think I'm using the fact that each rigid motion in $$D_n$$ extends uniquely to an isometry of $$\mathbb{R}^2$$, and I don't know if I'm answering Conrad's question "How are vertices unlike other points in terms of their distance relationships with all points on the polygon?".

• zipirovich’ answer shows that $g$ carries vertices to vertices. Your reasoning why $g$ preserves adjacency of vertices is correct. An application of Lemma 2.1 is not needed. May 19 '19 at 18:48

I think that maybe he hints at the following argument. For any point $$P$$ in such a regular polygon $$G$$ define $$d(P)=\max\limits_{Q\in G}\{\operatorname{dist}(P,Q)\}$$, the maximal possible distance between the given point and all other points in the polygon. The value of $$d(P)$$ is largest when $$P$$ is a vertex, and smaller for any other point in the polygon. And since rigid motions preserve distances, the value of $$d(P)$$ would be preserved for any point. Thus vertices have to be mapped to vertices, where the value is again largest possible.
• I'm not sure how can I prove that $d(P)$ is largest when $P$ is a vertex. Maybe another argument can work, too. A rigid motion in $D_n$ takes the polygon back to itself, so the circumscribed circle and its center remain the same. The vertices are the only points of the polygon whose distance from the center is equal to the radius of the circumscribed circle. Threfore, an element of $D_n$ must carry vertices to vertices. Is that correct? Jul 17 '19 at 15:41
My try: A boundary point of a convex $$n$$-gon $$K$$ has the special property that there exist sequences $$x_i \in K$$ and $$y_i \in \mathbb{R}^2\setminus K$$ that both converge to that point.
A vertex has the special property that there exists a straight line passing through the point, such that there are sequences on both rays, which converge to the vertex. This is a property that makes the vertices unique among the points in $$K$$.