The answer is rhombus only, if the polygon is concave say at (largest possible) contiguous concave stretch $A_1A_2 \cdots A_k$ reflect $A_i$ about $A_1A_k$, of all of them lie on a side then the stretch will not be concave contradiction ,now assume that the polygon is convex then the idea is to define ‘closest diagonal’ as the diagonal joining adjacent points, take a point $A$ for whom the distance between the closest diagonal and $A$ is minimised then take the vertex opposite to this diagonal (defined as the point having maximal distance from that diagonal), say $B$ then we claim that if the number of vertices are greater than $4$ then the reflection of $B$ over the diagonal does not lie on any side or inside, indeed assume to our contrary that it lies on boundary or interior then assume it lies on side then it is equivalent to saying that the distance from $B$ to the closest diagonal of $A$ is less than the distance of $A$, which is clearly false as the distance from $B$ to that diagonal is at least the distance from $B$ to its closest diagonal which is greater than the distance from $A$ to its closest diagonal bydefinition contradiction! and therefore $n=4$, in this case we must have a rhombus by repeating the argument.