Understanding the role of vertices in $D_{2n}$ in the definition of rigid motion I'm trying to understand a paragraph from Keith Conrad's expository paper on the dihedral group. The paragraph is:

Pick two adjacent vertices of a regular $n$-gon, and call them $A$ and $B$ as in the figure below. 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 distance relationships with all points in the polygon?) and $g$ must preserve adjacency of vertices, so $g(A)$ and $g(B)$ are adjacent vertices of the polygon.

The point I've bolded is what is causing me confusion. I can't figure out how vertices differ from other points. Is the idea that a symmetry has to send vertices to vertices, and adjacent vertices have to be mapped to adjacent vertices? I'm struggling to answer this in a way that ties back to the definition of a rigid motion.
Can someone help me to understand this?
 A: Although I  don't know what Conrad has in mind, one possible definition is this: a vertex of a set $P$ - in a Euclidean plane $E,$ say - is a point $v \in P$ that does not lie in any "open line segment" with endpoints in $P.$
This property can be defined in terms of the distance function $d \colon E \times E \to E,$ $(x, y) \mapsto \|x - y\|_2$ - itself usually defined in terms of the Euclidean norm function $\|\cdot\|_2$ - as follows.
If $x$ and $y$ are points of $E,$ the open line segment $]x, y[$ - this is a temporary, ad hoc technical term, with an equally temporary notation - is the set of all points $z \in E \smallsetminus\{x, y\}$ such that $d(x, z) + d(z, y) = d(x, y).$
Thus, a vertex of $P$ is defined as a point $v \in P$ such that there do not exist distinct points $x, y \in P \smallsetminus \{v\}$ such that $d(x, v) + d(v, y) = d(x, y).$
By definition (see the footnote on page 1 of Conrad's paper), a rigid motion is a function $T \colon E \to E$ - it is necessarily a bijection, whose inverse is also a rigid motion - such that $d(Tx, Ty) = d(x, y)$ for all $x, y \in E.$
Lemma. If $T$ is a rigid motion of $E,$ and $P$ is a subset of $E,$ then for all $v \in E,$ the point $v$ is a vertex of the set $P$ if and only if $Tv$ is a vertex of the image set $TP = \{Tx : x \in P\}.$
Proof. Because $T$ is a bijection, it is enough to prove the "if" part. Equivalently: if $v$ is not a vertex of $P,$ then $Tv$ is not a vertex of $TP.$ Suppose, then, that $v$ is not a vertex of $P.$ By definition, there exist distinct points $x, y \in P,$ neither equal to $v,$ such that $d(x, v) + d(v, y) = d(x, y).$ Because $T$ is a bijection, the points $Tv, Tx, Ty$ of $TP$ are pairwise distinct. Also, because $T$ is a rigid motion,
$$
d(Tx, Tv) + d(Tv, Ty) = d(x, v) + d(v, y) = d(x, y) = d(Tx, Ty).
$$
Therefore, $Tv$ is not a vertex of $TP.$ $\square$
It follows that if $T$ is a rigid motion of $E$ that maps a set $P$ to itself, i.e., $TP = P,$ then a point $v$ is a vertex of $P$ if and only if $Tv$ is a vertex of $P.$ In other words, $T$ restricts to a permutation of the set of vertices of $P.$
In the case where $n$ is an integer $\geqslant 3,$ and $P$ is a regular $n$-gon, it is clear that the set of vertices of $P,$ when defined in this way, is what we would expect. (But it might be a little tedious to write out a strict proof.) If $s$ is the common length of all the sides of $P,$ then two vertices $v, w$ of $P$ are said to be adjacent if and only if $d(v, w) = s.$ It is clear that if $T$ maps $P$ to itself, then $v, w$ are adjacent if and only if $Tv, Tw$ are adjacent.
A: Checkout this question: how not both points can be on the polygon implies that every point on the polygon is uniquely determined by its distance from given two points? (This fellow too was going through the same paper)
And yes, symmetry implies adjacent vertices being mapped to adjacent vertices.
