In circle $ O$, $ G$ is a moving point on diameter $ \overline{AB}$. $ \overline{AA'}$ is drawn perpendicular to $ \overline{AB}$ and equal to $ \overline{AG}$. $ \overline{BB'}$ is drawn perpendicular to $ \overline{AB}$, on the same side of diameter $ \overline{AB}$ as $ \overline{AA'}$, and equal to $ BG$. Let $ O'$ be the midpoint of $ \overline{A'B'}$. Then, as $ G$ moves from $ A$ to $ B$, point $ O'$:

$ \textbf{(A)}\ \text{moves on a straight line parallel to }{AB}\qquad \\
\textbf{(B)}\ \text{remains stationary}\qquad \\
\textbf{(C)}\ \text{moves on a straight line perpendicular to }{AB}\qquad \\
\textbf{(D)}\ \text{moves in a small circle intersecting the given circle}\qquad \\
\textbf{(E)}\ \text{follows a path which is neither a circle nor a straight line}$
1. **Define the problem setup:**
   - Let \( O \) be the center of the circle.
   - \( AB \) is the diameter of the circle.
   - \( G \) is a moving point on \( AB \).
   - \( AA' \) is perpendicular to \( AB \) and equal to \( AG \).
   - \( BB' \) is perpendicular to \( AB \) and equal to \( BG \).
   - \( O' \) is the midpoint of \( A'B' \).

2. **Analyze the positions of \( A' \) and \( B' \):**
   - Since \( AA' \) is perpendicular to \( AB \) and \( AA' = AG \), \( A' \) is a point on the circle centered at \( A \) with radius \( AG \).
   - Similarly, \( BB' \) is perpendicular to \( AB \) and \( BB' = BG \), so \( B' \) is a point on the circle centered at \( B \) with radius \( BG \).

3. **Determine the coordinates of \( A' \) and \( B' \):**
   - Assume \( A \) is at \( (-R, 0) \) and \( B \) is at \( (R, 0) \), where \( R \) is the radius of the circle.
   - Let \( G \) be at \( (x, 0) \) where \( -R \leq x \leq R \).
   - Then, \( A' \) is at \( (x, AG) = (x, x + R) \).
   - \( B' \) is at \( (x, BG) = (x, R - x) \).

4. **Find the midpoint \( O' \) of \( A'B' \):**
   - The coordinates of \( A' \) are \( (x, x + R) \).
   - The coordinates of \( B' \) are \( (x, R - x) \).
   - The midpoint \( O' \) of \( A'B' \) is:
     \[
     O' = \left( \frac{x + x}{2}, \frac{(x + R) + (R - x)}{2} \right) = \left( x, \frac{2R}{2} \right) = (x, R)
     \]

5. **Analyze the movement of \( O' \):**
   - As \( G \) moves from \( A \) to \( B \), \( x \) varies from \(-R\) to \(R\).
   - The \( x \)-coordinate of \( O' \) is \( x \), which varies from \(-R\) to \(R\).
   - The \( y \)-coordinate of \( O' \) is always \( R \).

6. **Conclusion:**
   - The point \( O' \) moves along a straight line parallel to \( AB \) at a distance \( R \) above \( AB \).

The final answer is \( \boxed{ \textbf{(A)}\ \text{moves on a straight line parallel to }{AB} } \)