Here are some rows trying to make (more) clear the situation described in the question. First of all, we need a handy notation, and clear definition of the action.
Some words on the described action first. It involves $D_n$ in the notation from Dihedral group, wiki page, the dihedral group of all plane symmetries that invariate the set of the vertices of a fixed $n$-gon. And these vertices, labelled by $1,2,\dots,n$. Let us denote by $[n]$ this set of $n$ vertices. Then the given (to-be-)action is a map first:
$$
D_n\times [n]\to [n]\ .
$$
If this is (soon) an action, then it is an action of the group $D_n$ on the set $[n]$. And there is no need to consider permutations of $[n]$ (and/or the group of permutation of $[n]$ and/or the image of $D_n$ inside this group of permutations).
Maybe the example in loc. cit. wants to do something with the permutation $\sigma_\alpha$ associated to some symmetry $\alpha$ of the plane containing the vertices of the $n$-gon. But for the question, we just do not need this notational complication.
The map above is defined by $(\alpha,i)\to\alpha\cdot i:=\alpha(i)$.
Here, $\alpha(i)$ is the result of applying the symmetry (= isometric transformation, which is in particular a function) $\alpha$ on the vertex $i$.
Let us check, that this leads to an action. For the identic map id (the neutral element in $D_n$), and for symmetries $\alpha,\beta$ and some vertex $i$ we have:
$$
\begin{aligned}
\operatorname{id}\cdot i
&=
\operatorname{id}(i)=i\ ,
\\[3mm]
(\alpha\circ\beta)\cdot i
&=(\alpha\circ\beta)(i)\\
&=\alpha(\beta(i))\\
&=\alpha\cdot(\beta(i))\\
&=\alpha\cdot(\beta\cdot i)\ .
\end{aligned}
$$
So we have indeed an action, and the above is all regarding a proof.
The action is faithful for $n>1$. (For $n\ge 3$ this follows from the fact that an affine plane is "determined by a triple of points", i.e. by fixing an origin and two unit vectors from the origin, a first one and a second one. For $n=2$ - just in case we really want to consider a "$2$-gon", the reflection w.r.t. the line through the two points is fixing pointwise the two vertices, so the action is not faithful. The case $n=1$ is also not coming with a faithful action.)
In case we really want to work with permutations, then the group of permutations acting on $[n]$ is the image $\sigma(D_n)$ of $D_n$ inside the permutation group of the set $[n]$ via the map $\alpha\to\sigma_\alpha$, and this action is defined by transport of structure, i.e. $\sigma_\alpha\cdot i :=\alpha\cdot i:=\alpha(i)$. (By transport, it is also an action on the set $[n]$.)
Best Answer
What Artin is saying is excactly that
is not true for $D_1$ and $D_2$ (at least with the definition he uses for that notation).
By the symmetry of the $n$-gon you should, in this case, think of the group of all the isometries of the plane that fix the outline of the $n$-gon. For $n=1$ there are two such isometries, namely the identity and the reflection through the midline.
It's a general fact about isometries of the plane that knowing what they do to three non-collinear points is enough to reconstruct the entire isometry. In particular for $n\ge 3$ it just happens that knowing what happens to each vertex of an $n$-gon is enough to reconstruct an entire isometry, so in that particular case one can represent elements of $D_n$ by permutations from $S_n$. But that's just a practical property that holds in the case $n\ge 3$, not part of the definition of $D_n$.