I've heard that rotational symmetry is where a figure is rotated a bit before it looks the same. in that case, what rotational symmetry is there on a circle? is there none, or is there infinite, or is it something else?
[Math] rotational symmetry on a circle
geometryrotations
Related Solutions
Let $X=(0,0)$ and let $W=(a,b)$. Call $S_X, S_W$ the respective squares, $S_X', S_W'$ after rotation.
$$S_X : (a\pm EF/2, b\pm EF/2)$$
$$S_W : (\pm AB/2, \pm AB/2)$$
After rotation:
$$S_X' : (a, b\pm \sqrt{2}/2EF),(a\pm \sqrt{2}/2EF, b)$$
$$S_W' : (0, \pm \sqrt{2}/2AB), (\pm \sqrt{2}/2AB,0)$$
(you can get this by multiplying the coordinates by the rotation matrix.)
If you want to line up $B$ and $F$ then you want real numbers $s,t$ such that
$$(s, \sqrt{2}/2AB+t)=(a, b+ \sqrt{2}/2EF)$$
So $$(s,t)= (a,b+\sqrt{2}/2(EF-AB))$$
Edit
Here is an interactive plot I made for you. First both squares of specified length are rotated by 45 degrees. Then the square centered at the origin is translated by the amount specified above.
For rectangles, rather than go through a whole 'nother derivation, the value of $s,t$ you seek is:
$$\left[\begin{matrix}s\\ t\end{matrix}\right]=\left[\begin{matrix}a\\ b\end{matrix}\right]+\left[\begin{matrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{matrix}\right]\left[\begin{matrix}F_x-B_x-a\\ F_y-B_y-b\end{matrix}\right]$$
Another example for rectangles, rotating $270^o$
Alpha symmetry is the largest angle a part would have to be turned about an axis perpendicular to the insertion axis.
Beta Symmetry is the largest angle the part would have to be rotated about the insertion axis for mating.
Alpha and Beta symmetry actually range from 0 to 360° (instead of the intuitive 0 to 180°) because it is assumed that the worst case rotation is used. reference Design.
Best Answer
Simple, there is an infinite order of rotation.
That is because the the centroid, the centre, has the same distance to any point on the circumference.
Common sense: it cannot be zero as there is more than no order of rotation.