Why do only the linear transformations that send an edge to an edge preserve the n-gon?

Question

I am asked to compute all the linear transformations $A:V\to V$ that preserve the general $n$-gon. We define the $n$-gon by using the roots of unity as vertices. I know that there are $2n$ such transformations, $n$ rotations and $n$ reflections, all of which send an edge to an edge.

A linear transformation is completely determined by its action on a basis; so, I take the vectors defined by two adjacent vertices, $v_1$ and $v_2$, as my basis, and define a linear transformation by choosing the points on the $n$-gon where I want to send them. Not every linear transformation that sends $v_1$ and $v_2$ to distinct vertices preserves the $n$-gon, though. Why do only the linear transformations that send $v_1$ and $v_2$ to adjacent vertices preserve the $n$-gon.

It might also be possible to approach this problem in another way because rotations are akin to multiplication in the complex field.

Answer 1

The edge $e$ from $v_1$ to $v_2$ can be parametrized as the set of points $(1-t)v_1 + tv_2$ for $0 \le t \le 1$. Since the transformation $A$ is linear, the image of $e$ under $A$ will be $(1-t)A(v_1) + tA(v_2)$ which is the edge from $A(v_1)$ to $A(v_2)$.

Answer 2

Hint Two (nonantipodal) vertices $v_1, v_2$ are adjacent iff there is no vertex $v'$ that can be written as a linear combination of $v_1, v_2$ with positive coefficients.

So, if $A v_1, A v_2$ are not adjacent (NB they cannot be antipodal), there is some vertex $A v'$ that we can write as $$A v' = a_1 A v_1 + a_2 A v_2 = A(a_1 v_1 + a_2 v_2)$$ for some positive $a_1, a_2$, and, since $A$ is invertible, $v' = a_1 v_1 + a_2 v_2$. Thus, $v_1, v_2$ are not adjacent.