Switzerland Selection Solution

Solution:

Write $P(x) = (x - z_{1})(x - z_{2}) \ldots (x - z_{2023})$, where $z_{1}, \ldots, z_{2023}$ are the roots of $P$. Note that $P(1) = 1$ is equivalent to $(1 - z_{1})(1 - z_{2}) \ldots (1 - z_{2023}) = 1$. Furthermore, by Vieta, $z_{1} + z_{2} + \ldots + z_{2023} = 0$. This gives us the two equations
$$\begin{aligned} (1 - z_{1}) + (1 - z_{2}) + \ldots + (1 - z_{2023}) & = 2023 \\ (1 - z_{1}) \cdot (1 - z_{2}) \cdot \ldots \cdot (1 - z_{2023}) & = 1 \end{aligned}$$
By observing that
$$\frac{1}{2023} \cdot \sum_{i=1}^{2023} (1 - z_{i}) = \left( \prod_{i=1}^{2023} (1 - z_{i}) \right)^{\frac{1}{2023}}$$
and using the equality case of AM-GM (which can be applied since $1 - z_{i} > 0$ for all $i$), we deduce that $z_{1} = z_{2} = \ldots = z_{2023} = 0$. Therefore, the only solution is $P(x) = x^{2023}$. This polynomial obviously satisfies the desired properties.