Let me try to give partial answer to this using a little bit of analysis along with linear algebra.
We are basically finding line of reflection which does not change shape of Polygon at all.
Now any reflection is a linear transformation(these are nothing but orthogonal matrices with deterimant -1). Now any linear transformation in $R^2$ is continuous and its inverse being same reflection also continuous, so it is homeomorphism.
So it fixes boundary of any set, takes convex set to convex and lines to lines(easy to prove).
Now corners are nothing but points on intersection of two line segment which lie on boundary and every other point on boundary is not intersection of two line segments lying on boundary.
Now under required reflection since line goes to line and intersection points of two line will go to intersection point of two line segments. Corner will go to corner.
Hence if $x_1, x_2,\dots, x_n$ corner points then $v= x_1,+x_2+\dots+x_n/n$ will remain invariant under given transformation. And it lies in our polygon being convex linear combination of points, infact it lies in the interior of polygon i.e. there exist a ball around it lying completely inside polygon.
Now under any reflection, only points on a line are fixed and every other point changes their position. Hence that line has to pass through $v$.
Now as point in interior of polygon, every line passing through it will intersect boundary.(shown on Convex Set and an interior point)
Claim: any line passing through $v$ can only intersect boundary at corner or at midpoints of sides(or line segment joining corners).
Proof: If it intersect at any other point, we can transform coordinates to make that point origin and line passing through $v$ as $x-$ axis. Now side which was intersected is now line segment passing through origin. Now it is easy to pove that any line segment if reflected alone $x-$ axis remain unaffected, then midpoint of line lies on origin
(use endpoint has to go to endpoint, and being reflection along x-axis), then $x-$ coordinates of endpoint will become zero and line segment would be just perpendicular to $x-$ axis. Now it is obvious that endpoint lies on origin.
Now using claim and fact since there are only $n$ corner points and $n$ sides, we get that line of symmetry can be atmost $2n$.
Now using result mentioned in Convex Set and an interior point
we also get that any line passing through $v$ will intersect boundary at atleast 2 points because $v$ is in interior of polygon.
And as any line passing thtough $v$ is determined by another point on it. Hence we get that under equivalence relation of being on same line of symmetry passing through $v$ on the set consisting of all corner and midpoints through which a symnetry line pass, we get that each equivalence class will have atleast 2 elements. And no. of equivalence class will be equal to no. of symmety lines.
We get since no. of equivalnece classes has to be less than equal to $n$ as set has atmost $2n$ elements while each equivalence class at atleast 2 elements.
We get line of symmetry can be atmost $n$
P.S. We proved for general n-polygon, there are atmost n line of symmetry. If for regular you want me to prove there are exactly $n$, that I don't know.
P.S. We also get by same argument $v$ is only point fixed under rotation, and as each vertex has to go to vertex, there are atmost $n$ rotations.