Midpoint polygons?

Note that the midpoints of a quadrilateral form a parallelogram. So if a quadrilateral is not parallelogram it cannot be the midpoint polygon of another quadrilateral.

EDIT:

Let the vertices of the polygon be of coordinates $(x_1,y_1)\ldots(x_n,y_n)$ Suppose that we start from $(a,b)$ and we incrementally construct the broken line $(a,b),(a_1,b_1)\ldots(a_n,b_n)$ having midpoints as the original polygon:

$x_1=\frac{a+a_1}{2}$, $y_1=\frac{b+b_1}{2}$ so

$a_1=-a+2x_1$, $b_1=-b+2y_1$

$a_2=-a_1+2x_2$, $b_2=-b_1+2y_2$

$\ldots$

$a_n=-a_{n-1}+2x_n$, $b_n=-b_{n-1}+2y_n$

To have a closed polygon one needs $a_n=a$ and $b_n=b$

So for $n$ odd you get an unique solution:

$a=x_1-x_2+x_3-\ldots+(-1)^{n+1}x_n$

$b=y_1-y_2+y_3-\ldots+(-1)^{n+1}y_n$

For $n$ even, the two linear systems are not of full rank (have either an infinity of zero solutions). You may check when each of the case happens with Rouche's theorem

So when $n$ is even, the system has solution iff:

$x_1-x_2+x_3-\ldots-x_n=0$

$y_1-y_2+y_3-\ldots-y_n=0$

In which case it has an infinity of solutions, starting with any point of the plane.

Of course depending on your unspecified requirements on the polygon (non-degenerate, non-self-intersect, convex), you might have to impose additional restrictions.

The midpoint transformation can be written in matrix form $Q=AP$, where $Q$ are the new vertices, $P$ are the old vertices, and the matrix $A$ is a circulant matrix whose first row is $1/2,1/2,0,\dots,0$.

The relevant property is that the rank of $A$ is $n-d$, where $d$ is the degree of $\gcd( x^{n-1}+1, x^n - 1)$. When $n$ is even, this gcd has at least the factor $x+1$ and so $A$ is singular and the midpoint transformation cannot be inverted.

The gcd is actually a divisor of $x+1= x(x^{n-1}+1)-(x^n - 1)$ and so is either $1$ or $x+1$. Therefore, $A$ is singular iff $n$ is even.