Just use complex numbers, with $ a = 1$, $ b = i$, $ c = - 1$ and $ d = - i$. With some calculations, we have $ k = \frac {\sqrt {3} - 1}{2}( - 1 - i)$, $ l = \frac {\sqrt {3} - 1}{2}(1 - i)$, $ m = \frac {\sqrt {3} - 1}{2}(1 + i)$ and $ n = \frac {\sqrt {3} - 1}{2}( - 1 + i)$. Now it's an easy job to calculate the twelve midpoints and to find out they are all of the form $ \frac {\sqrt {3} - 1}{2}e^{\frac {k\pi}{6}i}$, with $ k\in\mathbb{N}: 0\le k\le 11$, and the result follows.

Synthetic solution: Clearly the four first midpoints form a square with midpoint $O$ where $O$ is the midpoint of $ABCD$. Hence it suffices to check that the other eight points have the same distance from $O$ and that the angles at $O$ are correct. By symmetry, it will then suffice to check that $OEF$ is equilateral where $E$ is the midpoint of $KL$ and $F$ is the midpoint of $AK$. Being midlines in triangles $AKL$ and $AKC$, we see that $EF=\frac{1}{2} AL$ and $OF=\frac{1}{2} KC$. On the other hand, in the square $KLMN$ we clearly have $EO=\frac{1}{2} KL$. But by construction, the triangles $ABL, LBK, KBC$ are all congruent whence $AL=LK=KC$ and hence $EF=EO=OF$ as desired. [asy][asy] import olympiad; size(6cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.4)); dotfactor *= 1.5; pair A = dir(225), B = dir(315), C = dir(45), D=dir(135), K=extension(A,dir(75),B,dir(105)), L=extension(B,dir(165), C,dir(195)), M=-K, N=-L, E=(K+L)/2, F=(A+K)/2, O=0; draw(A--B--C--D--A); draw(B--K--A, dotted); draw(B--L--C, dotted); draw(K--L, dashed); draw(E--F--O--E, dashed); draw(A--L, dashed); draw(C--K, dashed); dot("$A$", A, dir(225)); dot("$B$", B, dir(315)); dot("$C$", C, dir(45)); dot("$D$", D, dir(135)); dot("$K$", K, dir(90)); dot("$L$", L, dir(180)); dot("$M$", M, dir(270)); dot("$N$", N, dir(0)); dot("$O$", O, dir(0)); dot("$E$", E, dir(135)); dot("$F$", F, dir(315)); dot((M+L)/2); dot((M+N)/2); dot((N+K)/2); dot((B+K)/2); dot((B+L)/2); dot((C+L)/2); dot((C+M)/2); dot((D+M)/2); dot((D+N)/2); dot((A+N)/2); [/asy][/asy]