1. Given a natural number $m \geqslant 2$, find all polynomials that satisfy $P[P(x)]=P^{m}(x)$.
1. If $\operatorname{deg} P=0$, let $P(x)=a$, substitute into the given equation, we get $a=a^{m}$. Since $m>1$, we have $a=0$ or $a^{m-1}=1$. Therefore, $a=0$ or $a=e^{\frac{2 k \pi i}{m-1}}, k=0,1, \cdots, m-2$.

If $\operatorname{deg} P>0$, take any complex number $\alpha$, by the fundamental theorem of algebra, the equation $P(x)=\alpha$ must have a root $\beta$, i.e., $P(\beta)=\alpha$. By the given condition, we have $P[P(\beta)]=P^{m}(\beta)$, i.e., $P(\alpha)=\alpha^{m}$. By the arbitrariness of $\alpha$ and the identity theorem of polynomials, we get $P(x)=x^{m}$.