Is there a polynomial $g(x)$ such that $p \circ f=g \circ p$, where $f(x)=9x+30x^3+27x^5+9x^7+{{x}^{9}}$? Let us consider a polynomial $f(x)=9x+30x^3+27x^5+9x^7+{{x}^{9}}$ over a field of characteristic $0$.
Set $p(x)=x^nh(x), n \geq 1$, where $h(x)$ can be any rational function.
I am trying to find out another polynomials $g(x)$ such that $p \circ f=g \circ p$. One more information, for any zero $\alpha$ of $f(x)$, $~~p(\alpha)$ is a root of $g(x)$.
If $h(x)=1$, then we need find a $g(x)$ such that $(f(x))^n=g(x^n)$.
I exclude the case $f = g, n=1$.
I have found the roots of $f(x)$ here in decimal form.
I thing this gives a hints about $g(x)$ because $f(\alpha)=0 \Rightarrow f(p(\alpha))=0$.
Edit: sorry for multiple edits but I hope this is the final one

$$
\begin{array}{rcl}
p(x)&=&9x,\\
g(x)&=&\displaystyle \frac{x^9}{43046721}+\frac{x^7}{59049}+\frac{x^5}{243}+\frac{10 x^3}{27}+9 x.
\end{array}
$$
Upd. For the new version
$$
\begin{array}{rcl}
p(x)&=&x^2,\\
g(x)&=&x (x+3)^2 \left(x (x+3)^2+3\right)^2;
\end{array}
$$
$$
\begin{array}{rcl}
p(x)&=&x^4+4 x^2,\\
g(x)&=&x (x+3)^2 \left(x (x+3)^2+3\right)^2.
\end{array}
$$
$$
\begin{array}{rcl}
p(x)&=&x^2 \left(x^2+3\right)^2,\\
g(x)&=&x (x+3)^2 \left(x (x+3)^2+3\right)^2.
\end{array}
$$
