For every $n\geq 3$, determine all the configurations of $n$ distinct points $X_1,X_2,\ldots,X_n$ in the plane, with the property that for any pair of distinct points $X_i$, $X_j$ there exists a permutation $\sigma$ of the integers $\{1,\ldots,n\}$, such that $\textrm{d}(X_i,X_k) = \textrm{d}(X_j,X_{\sigma(k)})$ for all $1\leq k \leq n$.
(We write $\textrm{d}(X,Y)$ to denote the distance between points $X$ and $Y$.)

[i](United Kingdom) Luke Betts[/i]
1. **Understanding the Problem:**
   We need to determine all configurations of \( n \) distinct points \( X_1, X_2, \ldots, X_n \) in the plane such that for any pair of distinct points \( X_i, X_j \), there exists a permutation \( \sigma \) of the integers \( \{1, \ldots, n\} \) such that \( \textrm{d}(X_i, X_k) = \textrm{d}(X_j, X_{\sigma(k)}) \) for all \( 1 \leq k \leq n \).

2. **Initial Observations:**
   The condition implies that for any \( i, j \), the multisets \( \{ \textrm{d}(X_i, X_k) \}_k \) and \( \{ \textrm{d}(X_j, X_k) \}_k \) are the same. This means that the set of distances from any point \( X_i \) to all other points is the same for all \( i \).

3. **Using the Centroid:**
   Let \( G \) be the centroid of the points. By the parallel axis theorem, we have:
   \[
   \sum_{k=1}^n \textrm{d}(X_i, X_k)^2 = n \cdot \textrm{d}(G, X_i)^2 + \sum_{k=1}^n \textrm{d}(G, X_k)^2
   \]
   Since the left-hand side is the same for all \( i \), it follows that \( \textrm{d}(G, X_i) \) is constant for all \( i \). This implies that all points \( X_i \) lie on a circle centered at \( G \).

4. **Relabeling and Convexity:**
   Without loss of generality, relabel the points such that \( X_1, X_2, \ldots, X_n \) form a convex polygon. Let \( \Omega \) be the circumcenter of this polygon.

5. **Analyzing the Polygon:**
   - The polygon must contain the center of \( \Omega \), otherwise, the edge closest to the center would be strictly longer than any other distance between points, which contradicts the given condition.
   - Consider the shortest edge in \( X_1, \ldots, X_n \), and assume it is \( X_1X_2 \). For any \( i \neq j \), we have \( \textrm{d}(X_i, X_j) \geq \min\{\textrm{d}(X_i, X_{i-1}), \textrm{d}(X_i, X_{i+1})\} \) since \( X_j \) does not lie on the arc \( X_{i-1}X_iX_{i+1} \).

6. **Forcing Alternating Side Lengths:**
   - For each \( i \), we must have \( X_1X_2 \in \{\textrm{d}(X_i, X_{i-1}), \textrm{d}(X_i, X_{i+1})\} \).
   - If for some \( i \), \( \textrm{d}(X_i, X_{i-1}) = \textrm{d}(X_i, X_{i+1}) = X_1X_2 \), then by a similar argument, we force \( \textrm{d}(X_j, X_{j-1}) = \textrm{d}(X_j, X_{j+1}) = X_1X_2 \) for all \( j \).

7. **Conclusion for Odd and Even \( n \):**
   - For \( n \) odd, this forces all sides to be equal, making the polygon regular.
   - For \( n \) even, this is also forced unless we have \( X_1X_2 = X_3X_4 = \cdots = X_{n-1}X_n \). In this case, the sides must alternate in length.

Thus, the configurations are:
- For \( n \) odd: The points form a regular \( n \)-gon.
- For \( n \) even: The points form a convex cyclic \( n \)-gon with alternating side lengths.

\(\blacksquare\)

The final answer is: For \( \boxed{ n } \) odd, the points form a regular \( n \)-gon. For \( n \) even, the points form a convex cyclic \( n \)-gon with alternating side lengths.