Can you fix uncountably many holes in a circle using one rotation? Let's say you have a circle with a single hole in it (it is exactly the same as a circle, except there is exactly one point a complete circle has that this figure does not). There is a way to fix this circle's hole that does not make any new holes with only one rotation.
Let's say the circle starts red. Go $\sqrt 2^\circ$ to the right of the hole, and color it green. Now go $\sqrt 2^\circ$ to the right of that point and color it green. Continuing circling around $\sqrt 2^\circ$ coloring points green ad infinium. Now take all the green points, and rotate them  $\sqrt 2^\circ$ left. The hole will be filled by the first green point, the first green points hole by the second green point, the second by the third, and so on. Since for every green point there is a next green point, no green point becomes a hole. We have fixed the hole with one rotation!
The reason why $\sqrt 2 ^ \circ$ was chosen is so that when we went to color a green point, we did not end up coloring the hole instead (which would have made the above procedure incorrect). If the hole was the $n$th green point, and we had gone around the circle $m$ times, then $n*\sqrt 2=360*m$. But this implies $\sqrt 2 = \frac{360m}n$, a contradiction by $\sqrt 2$'s irrationality.
This construction works with more holes.  Indeed, say we have a countable set $H$ of holes. We need to find an angle $\theta$ such that a hole will need be where the green point of another will be. Consider the following: choose a point $h_1 \in H$, and $h_2 \in H$, a clockwise angle $\phi$ from $h_1$ to $h_2$, and a natural number $n\gt0$. If $\theta*n=\phi$, then $n$ green points after $h_1$, it will fail since $h_2$ is missing from the circle. Notice that this is the only way these scheme can fail. If you choose a $\theta$ such that no $h_1$, $h_2$, $\phi$, and $n$ exist to satisfy $\theta*n=\phi$, the scheme will work.
But $h_1$, $h_2$, $\phi$, and $n$ were all chosen from countable sets, and  $\aleph_0 *\aleph_0 *\aleph_0 *\aleph_0 = \aleph_0$. Each of these choices eliminates one angle ($\frac\phi n$), so $\aleph_0$ choices have been eliminated. But there are $\aleph_1$ real numbers from 0 to 360, so we can choose a $\theta$ that was not eliminated.
This means even with any countably infinite set of holes, we select some points, and rotate them all together to plug the holes without leaving any new ones.

My question is, when does this work when $H$ is uncountable? Can you fix uncountable holes? So far, I have not figured out how to do it for any uncountable set.
There are obviously some impossible examples. You can't fix a circle if you remove all the holes. You need to leave at-least uncountable points. This is not sufficient though: if you remove all the circle except a $60^\circ$ arc, you can't fill in the rest of the circle with one rotation.
In fact, even leaving $359^ \circ$ of the circle won't work, since every angle $\theta$ will be of the form $\frac \phi n$ at least once.
Is there any uncountable $H$ such that you can fix the holes? If you which $H$?
 A: Given a set of holes $H$, you want an angle $\theta$  such that $H$ and the rotations $R_{n\theta}(H)$, $n \in \mathbb N$, are all disjoint (where $R_\theta$ is rotation by $\theta$).  It's convenient to represent the circle as the unit circle $\mathbb T$ in the complex plane, so that $R_\theta$ is multiplication by $e^{i\theta}$. 
Suppose $H \subset \mathbb T$.  Then $R_\theta(H) \cap H = \emptyset$ iff $e^{i\theta} \notin H H^{-1} = \{st^{-1}: s,t \in H\}$.  To be able to fill the hole, you want $e^{in \theta} \notin H H^{-1}$ for all positive integers $n$, i.e. $e^{i\theta} \notin \bigcup_{n =1}^\infty (H H^{-1})^{1/n}$ (where $A^{1/n}$ consists of all the $n$'th roots of all members of $A$).  For such $\theta$ to exist, it is sufficient that $H H^{-1}$ has Lebesgue measure $0$.  In particular, this will be true if
the Hausdorff and upper box counting dimensions of $H$ are both less than $1/2$.  In particular, appropriate Cantor-type sets will have this property.
For a concrete example, let $H$ be the set of $e^{2 \pi i x}$ where $0 \le x < 1$ and
every decimal digit of $x$ is $0$ or $1$.  Then every member of $H H^{-1}$ can be written as $e^{2\pi i y}$ where every decimal digit of $y$ is 
$0,1,8$ or $9$.  
A: Fix a copy of the Cantor set on the circle. This is an uncountable set, whose complement is the union of countably many open intervals. Each open interval you can "push" towards its end whilst not moving any of the points on the Cantor set.
