jmc

By Vieta's formulas, the average of the sum of the roots is $\frac{6}{4}=\frac{3}{2},$ which corresponds to the center of the parallelogram. So, to shift the center of the parallelogram to the origin, let $w=z-\frac{3}{2}.$ Then $z=w+\frac{3}{2},$ so $${\left(w+\frac{3}{2}\right)}^4-6{\left(w+\frac{3}{2}\right)}^3+11a{\left(w+\frac{3}{2}\right)}^2-3(2a^2+3a-3)\left(w+\frac{3}{2}\right)+1=0.$$Hence, $$(2w+3{)}^4-2\cdot 6(2w+3{)}^3+4\cdot 11a(2w+3{)}^2-8\cdot 3(2a^2+3a-3)(2w+3)+16=0.$$Expanding, we get $$16w^4+(176a-216)w^2+(-96a^2+384a-288)w-144a^2+180a-11=0.$$The roots of this equation will form a parallelogram centered at the origin, which means they are of the form $w_1,$ $-w_1,$ $w_2,$ $-w_2.$ Thus, we can also write the equation as $$(w-w_1)(w+w_1)(w-w_2)(w+w_2)=(w^2-w_1^2)(w^2-w_2^2)=0.$$Note that the coefficient of $w$ will be 0, so $$-96a^2+384a-288=0.$$This equation factors as $-96(a-1)(a-3)=0,$ so $a=1$ or $a=3.$

For $a=1,$ the equation becomes $$16w^4-40w^2+25=(4w^2-5{)}^2=0,$$which has two double roots.

For $a=3,$ the given equation becomes $$w^4+312w^2-767=0.$$The roots of $x^2+312x-767=0$ are real, and one is positive and the other is negative. This mean that two of the roots of $w^4+312w^2-767=0$ are real (and negatives of each other), and the other two are imaginary (and negatives of each other), so they form a parallelogram.

Thus, the only such value of $a$ is $\boxed{3}.$