Let $n\geq 2$ be an integer. Determine all sets of real numbers $(a_1, a_2,\dots, a_n)$ such that $a_1-2a_2, a_2-2a_3,\dots, a_{n-1}-2a_n, a_n-2a_1$ are rearrangements of $a_1, a_2,\dots, a_n$. 
1. We are given that the set $\{a_1 - 2a_2, a_2 - 2a_3, \ldots, a_{n-1} - 2a_n, a_n - 2a_1\}$ is a rearrangement of $\{a_1, a_2, \ldots, a_n\}$. This implies that the sum of the elements in both sets must be equal:
   \[
   \sum_{k=1}^n a_k = \sum_{k=1}^n (a_k - 2a_{k+1})
   \]
   where indices are taken modulo $n$.

2. Simplifying the right-hand side, we get:
   \[
   \sum_{k=1}^n (a_k - 2a_{k+1}) = \sum_{k=1}^n a_k - 2\sum_{k=1}^n a_{k+1}
   \]
   Since the indices are taken modulo $n$, $\sum_{k=1}^n a_{k+1} = \sum_{k=1}^n a_k$. Therefore:
   \[
   \sum_{k=1}^n a_k - 2\sum_{k=1}^n a_k = \sum_{k=1}^n a_k - 2\sum_{k=1}^n a_k = -\sum_{k=1}^n a_k
   \]

3. Equating the sums from both sides, we get:
   \[
   \sum_{k=1}^n a_k = -\sum_{k=1}^n a_k
   \]
   Adding $\sum_{k=1}^n a_k$ to both sides, we obtain:
   \[
   2\sum_{k=1}^n a_k = 0 \implies \sum_{k=1}^n a_k = 0
   \]

4. Next, we consider the given condition in terms of the magnitudes of the complex numbers. We know:
   \[
   \sum_{k=1}^n |a_k|^2 = \sum_{k=1}^n |a_k - 2a_{k+1}|^2
   \]

5. Let $a_k = x_k + iy_k$ where $x_k, y_k \in \mathbb{R}$. Then:
   \[
   |a_k|^2 = x_k^2 + y_k^2
   \]
   and
   \[
   |a_k - 2a_{k+1}|^2 = (x_k - 2x_{k+1})^2 + (y_k - 2y_{k+1})^2
   \]

6. Expanding the right-hand side, we get:
   \[
   \sum_{k=1}^n \left( (x_k - 2x_{k+1})^2 + (y_k - 2y_{k+1})^2 \right)
   \]
   \[
   = \sum_{k=1}^n \left( x_k^2 - 4x_kx_{k+1} + 4x_{k+1}^2 + y_k^2 - 4y_ky_{k+1} + 4y_{k+1}^2 \right)
   \]
   \[
   = \sum_{k=1}^n \left( x_k^2 + y_k^2 \right) + \sum_{k=1}^n \left( 4x_{k+1}^2 + 4y_{k+1}^2 \right) - 4\sum_{k=1}^n \left( x_kx_{k+1} + y_ky_{k+1} \right)
   \]

7. Since $\sum_{k=1}^n |a_k|^2 = \sum_{k=1}^n |a_k - 2a_{k+1}|^2$, we have:
   \[
   \sum_{k=1}^n \left( x_k^2 + y_k^2 \right) = \sum_{k=1}^n \left( x_k^2 + y_k^2 \right) + 4\sum_{k=1}^n \left( x_{k+1}^2 + y_{k+1}^2 \right) - 4\sum_{k=1}^n \left( x_kx_{k+1} + y_ky_{k+1} \right)
   \]

8. Simplifying, we get:
   \[
   0 = 4\sum_{k=1}^n \left( x_{k+1}^2 + y_{k+1}^2 \right) - 4\sum_{k=1}^n \left( x_kx_{k+1} + y_ky_{k+1} \right)
   \]
   \[
   \sum_{k=1}^n \left( x_{k+1}^2 + y_{k+1}^2 \right) = \sum_{k=1}^n \left( x_kx_{k+1} + y_ky_{k+1} \right)
   \]

9. Using the Cauchy-Schwarz inequality, we know:
   \[
   \sum_{k=1}^n \left( x_k^2 + y_k^2 \right) \geq \sum_{k=1}^n \left( x_kx_{k+1} + y_ky_{k+1} \right)
   \]
   Since equality holds, it must be that $x_k = x_{k+1}$ and $y_k = y_{k+1}$ for all $k$. Therefore, $a_k = a_{k+1}$ for all $k$.

10. Given that $\sum_{k=1}^n a_k = 0$ and $a_k = a_{k+1}$ for all $k$, it follows that $a_1 = a_2 = \cdots = a_n = 0$.

Therefore, the only solution is:
\[
\boxed{a_1 = a_2 = \cdots = a_n = 0}
\]