How to find all polynomials that follows the equation $P(P'(x)) = P'(P(x))$ I want to find all polynomials that has the following property
$$P(P'(x)) = P'(P(x))$$
where P(x) is a polynomial.
Can you please tell me a way to solve this problem?

Let $ax^{n}$ be the leading coefficient, then the leading coefficient of both sides of the equation are equal giving:
$a(anx^{n-1})^{n} = an(ax^{n})^{n-1}$
So $a^{n+1}n^{n} = a^{n}n$
So $a= \frac{1}{n^{n-1}}$
So one such family is $\frac{x^{n}}{n^{n-1}}$
Now consider the constant coefficient K and the linear coefficient L:
$P'(K) = P(L)$
So if $P(x) = Lx+K$ is linear then, LK=K
So linear functions of the form
$P(x) = x+k$
Work.
$P(x) = ax^{2}+Lx+K$
Has $a=\frac{1}{2}$ and $\frac{L^{2}}{2}+L^{2} + K = K + L$
So K is free, and Either L=0 or $L = \frac{2}{3}$.
$L = \frac{2}{3}$ doesn't work when plugging back into $P'(P(x)) = P(P'(x))$
So we get $\frac{x^{2}}{2}+k$.
At third order and higher P and P' are both non-linear which forces K to be a solution of a polynomial of order higher than 2 and so K can only take on so many values. It is likely that the first family given will describe all of the remaining polynomials. We will demonstrate for the third order polynomials.
$\frac{x^{3}}{9} + bx^{2} + Lx + K$
Looking at 5th order terms:
$\frac{b}{27} = \frac{2b}{81}$
Which forces $b=0$.
Then looking at second order terms:
$\frac{L^{2}}{3} = 0$
which forces $L=0$.
$$\frac{K^{2}}{3} + 2bK + L = \frac{L^{3}}{9} + bL^{2} + L^{2} + K$$
So $\frac{K^{2}}{3} - K = 0$.
So K=3 or K=0.
$\frac{1}{3}(\frac{x^{3}}{9} + K )^2 = \frac{1}{9}(\frac{x^{2}}{3})^3 + K$
Looking at the 3rd order term forces K=0.
So $\frac{x^{3}}{9}$ is the only 3rd order solution.
