Find all polynomials : $ P(x^2-x)=xP(x-1)$

You can go also with canonical form of polynomial: $p(x) =a_nx^n+...$, where $a_n\ne 0$. So we have $$a_n(x^2-x)^n+...= xa_n(x-1)^n+...$$ so $a_nx^{2n}=a_nx^{n+1}$. Since $a_n\ne 0$ we have $n=1$ or $n=- \infty $. So $p(x)=ax+b$ for some $a,b$. Pluging in to starting equation we have:

$$a(x^2-x)+b= x(a(x-1)+b)\Longrightarrow ax^2-ax+b = ax^2+(-a+b)x$$

So $p(x)=ax$ is solution for any $a$.

The answer is fine but it can be made more precise in the same direction.

$$Q(x^2-x)=Q(x-1),x\neq0,1$$

Write this by substitution $y=x-1$ as

$$Q((y+1)y)=Q(y),y\neq0,-1$$

Now let us define the recursion

$$y_n=(y_{n-1}+1)y_{n-1},\ y_1=1$$

Obviously

$$y_n \gt y_{n-1}$$

$$Q(y_{n})=Q((y_{n-1}+1)y_{n-1})=Q(y_{n-1})=Q(1)$$

for any $n$, leading to your conclusion.