I haven't spent any time on the subsidiary Q, but as to the problem of an isometry $f$ of the polygon $P$ to itself, observe that the isometric image $f(S)$ of a closed bounded line segment $S$ of positive length $l$ must be a closed bounded line segment of equal length, because
(1)..... For any $\{x_1,x_2,x_3\}\subset S$ and for some $\{i,j,k\}=\{1,2,3\}$ we have $d(x_i,x_j)+d(x_j,x_k)=d(x_i,x_k)$ and hence $d(f(x_i),f(x_j))+d(f(x_j),f(x_k))=d(f(x_i),f(x_k)).$..... Therefore
($\bullet$) Any 3 points in $f( S)$ are collinear.
(2)..... Let $e_1,e_2$ be the endpoints of $S$. Let $T$ be the closed line segment from $f(e_1)$ to $f(e_2).$ Let $x$ be any point of $S$. Then $l= d(e_1,e_2)=d(f(e_1),f(e_2)),$ while $d(f(x),f(e_1))+ d(f(x),f(e_2))=d(x,e_1)+d(x,e_2)=l$ ...... So by ($\bullet$), we have
($\bullet \bullet)$ $f(S) \subset T.$ And $f(S)$ includes the endpoints of $T,$ and the length of $T$ is $l$.
(3)..... For any $y\in T ,$ there exists $x\in S$ with $d(x,e_i)=d(y,f(e_i))$ for $i\in \{1,2\},$ and by ($\bullet \bullet$) we have $f(x)\in T.$ So we must have $f(x)=y.$..... Hence $f(S)\supset T.$ Together with $f(S)\subset T$ from (2), we have $$ f(S)=T.$$
For the polygon $P,$ and an isometry $f:P\to P$, let $S$ be any of its sides. Then $f(S)$ is a closed straight-line segment of equal length and $f(S)\subset P,$ so $f(S)$ is a side of $P.$ And from ($\bullet \bullet$), $f$ maps the endpoints of $S,$ which are vertices of $P,$ to the endpoints of $f(S),$ which are also vertices of $P$, as they are the endpoints of the side $f(S). $