jmc

If we let $x=$ the side length of the triangle, then we can find the area of the triangle in terms of $x$ and then set it equal to $16\sqrt{3}$ to find $x$. The base of the triangle has length $x$. To find the altitude, we notice that drawing an altitude splits the equilateral triangle into two $30-60-90$ triangles with the longest side having length $x$. Since the ratio of the side lengths of a $30-60-90$ triangle is $1:\sqrt{3}:2$, the altitude will have length $\frac{x\sqrt{3}}{2}$ and the area of the triangle will be $\frac{1}{2}x\left(\frac{x\sqrt{3}}{2}\right)=\frac{x^2\sqrt{3}}{4}$. Setting this equal to $16\sqrt{3}$, we have that $\frac{x^2\sqrt{3}}{4}=16\sqrt{3}.$

Solving for $x$, we get that $x=8$. Since the side length of the triangle is $8$ and the square and triangle have equal perimeters, the square has a side length of $\frac{8\cdot 3}{4}=6$. If we draw the diagonal of the square, we notice that it splits the square into two $45-45-90$ triangles with legs of length $6$. A $45-45-90$ triangle has side length ratios of $1:1:\sqrt{2}$, so the diagonal of the square has length $\boxed{6\sqrt{2}}$ cm.