Solution:

Let $P$ be a polygon with $2n + 1$ sides. Let the vertices of $P$ be $A_1, A_2, \ldots, A_{2n+1}$ in order. The process consists of forming a new polygon $P'$ whose vertices are the midpoints $M_1, M_2, \ldots, M_{2n+1}$ of the sides $A_1A_2, A_2A_3, \ldots, A_{2n+1}A_1$ respectively. This process is repeated.

Let us represent the vertices of $P$ as complex numbers $z_1, z_2, \ldots, z_{2n+1}$ on the complex plane. The midpoints are then:

$$M_k = \frac{z_k + z_{k+1}}{2}, \quad k = 1, 2, \ldots, 2n+1$$
(with $z_{2n+2} = z_1$).

Let $T$ be the transformation that sends $(z_1, z_2, \ldots, z_{2n+1})$ to $(M_1, M_2, \ldots, M_{2n+1})$.

Let us consider the effect of repeatedly applying $T$.

Let $\omega = e^{2\pi i/(2n+1)}$ be a primitive $(2n+1)$-th root of unity. The set of vectors $(z_1, z_2, \ldots, z_{2n+1})$ can be expanded in the basis of $(1, \omega^k, \omega^{2k}, \ldots, \omega^{(2n)k})$ for $k = 0, 1, \ldots, 2n$.

The transformation $T$ acts linearly:

$$M_k = \frac{z_k + z_{k+1}}{2}$$

If we consider the vector $(z_1, z_2, \ldots, z_{2n+1})$ as a column vector $Z$, then $T$ is multiplication by a circulant matrix $A$ whose first row is $(1/2, 1/2, 0, \ldots, 0)$.

The eigenvectors of a circulant matrix are the vectors $(1, \omega^k, \omega^{2k}, \ldots, \omega^{(2n)k})$, and the corresponding eigenvalues are:

$$\lambda_k = \frac{1}{2} + \frac{1}{2} \omega^k$$
for $k = 0, 1, \ldots, 2n$.

Now, $\omega^k$ for $k = 1, \ldots, 2n$ are all distinct roots of unity except $1$.

The eigenvalue for $k = 0$ is $1$.

The modulus of $\lambda_k$ for $k \neq 0$ is:

$$|\lambda_k| = \left| \frac{1}{2} + \frac{1}{2} \omega^k \right| = \left| e^{i\theta_k/2} \cos(\theta_k/2) \right| = |\cos(\theta_k/2)|$$
where $\omega^k = e^{i\theta_k}$.

Since $2n+1$ is odd, none of the $\omega^k$ (for $k \neq 0$) is $-1$, so $\cos(\theta_k/2) \neq 0$ for all $k$.

But $|\lambda_k| < 1$ for all $k \neq 0$.

Therefore, as we repeat the process, all components except the $k=0$ component (the centroid) decay to zero. Thus, the sequence of polygons converges to a regular $(2n+1)$-gon centered at the centroid of $P$, which is homothetic to $P$.

Therefore, after sufficiently many steps, the polygon becomes homothetic to $P$.