Let the order of $x$ to be the minimum value $k$ such that $P^k(x)=x$ if it exists, otherwise it is $\infty$. Then, we claim that the only possible orders are $1$, $2$, and $\infty$. Suppose that there exists $x_1$, $x_2$, $\ldots$, $x_i$ such that $i\geq3$, $x_j$ are distinct, and $P(x_j)=x_{j+1}$ where $x_{i+a}=x_a$. Then, we have $x_j-x_{j+1}\mid x_{j+1}-x_{j+2}$ for all $j$. Multiplying all of these equations, we get $(x_1-x_2)(x_2-x_3)\ldots(x_i-x_1)\mid(x_1-x_2)\ldots(x_i-x_1)$. Therefore, this means that $|x_j-x_{j+1}|=|x_{j+1}-x_{j+2}|$ for all $j$. Suppose that $x_j-x_{j+1}=x_{j+2}-x_{j+1}$. Then, $x_j=x_{j+2}$, which contradicts the fact that $i\geq3$. Therefore, $x_j-x_{j+1}=x_{j+1}-x_{j+2}=c$ for some constant $c$. However, adding up all of these equations, we get $ic=0$, which implies $c=0$, so $x_j=x_{j+1}$, which is also a contradiction. Therefore, there does not exist a polynomial for $n\geq3$ since $n$ is not a possible order of any $x$.

Since $x-y|P(x)-P(y)$, $b_2-b_1|P(b_2)-P(b_1)=b_3-b_2|b_4-b_3|\cdots |b_n-b_{n-1}|b_1-b_n$, which implies $|b_2-b_1|=|b_3-b_2|=\cdots=|b_1-b_n|$. Consider (one of) the smallest $b_k$, wlog with $k=2$. It follows that $b_1=b_3$ (we couldn't have $b_2-b_1=b_3-b_2$ since this would imply that $b_2$ isn't the smallest), and so we have a contradiction since $n\geq 3$ and $b_1\neq b_3$