A circle is divided into $13$ segments, numbered consecutively from $1$ to $13$. Five fleas called $A,B,C,D$ and $E$ are sitting in the segments $1,2,3,4$ and $5$. A flea is allowed to jump to an empty segment five positions away in either direction around the circle. Only one flea jumps at the same time, and two fleas cannot be in the same segment. After some jumps, the fleas are back in the segments $1,2,3,4,5$, but possibly in some other order than they started. Which orders are possible ?
1. **Re-numbering the Circle:**
   We start by re-numbering the circle to simplify the problem. The original circle has segments numbered from $1$ to $13$. We re-number the segments such that each movement of $5$ steps in the original circle corresponds to $1$ step in the new circle. The new numbering is as follows:
   \[
   1, 6, 11, 3, 8, 13, 5, 10, 2, 7, 12, 4, 9
   \]
   This re-numbering is done in a counterclockwise direction.

2. **Initial Positions:**
   The initial positions of the fleas $A, B, C, D, E$ are in segments $1, 2, 3, 4, 5$ respectively. In the new numbering, these positions correspond to:
   \[
   1, 10, 3, 12, 5
   \]

3. **Allowed Jumps:**
   A flea can jump to an empty segment $5$ positions away in either direction. In the new numbering, this translates to moving $1$ step in either direction. Therefore, a flea in segment $i$ can move to segment $i+1$ or $i-1$ (modulo $13$).

4. **Permutations:**
   Since no two fleas can occupy the same segment, and each flea can only move to adjacent segments in the new numbering, the only possible permutations are cyclic permutations. This means that the fleas can only rotate their positions among the segments $1, 10, 3, 12, 5$.

5. **Cyclic Permutations:**
   The cyclic permutations of the positions $1, 10, 3, 12, 5$ are:
   \[
   (1, 10, 3, 12, 5), (10, 3, 12, 5, 1), (3, 12, 5, 1, 10), (12, 5, 1, 10, 3), (5, 1, 10, 3, 12)
   \]

6. **Conclusion:**
   Therefore, the only possible orders of the fleas after some jumps are the cyclic permutations of their initial positions.

The final answer is \( \boxed{ (1, 10, 3, 12, 5), (10, 3, 12, 5, 1), (3, 12, 5, 1, 10), (12, 5, 1, 10, 3), (5, 1, 10, 3, 12) } \)