Solve $f(x)f(2x^2) = f(2x^3+x)$

As observed in other answers, the leading coefficient of $f$, and the constant term both have to equal $1$ (if $f$ is not identically $0$). Now assume that $f$ has complex root with absolute value greater than 1. Let $z$ one of them with greatest absolute value. Now $2z^3 + z$ is also a root, by triangle inequality $$ |2z^3 + z| = |z||2z^2 + 1| \geq |z|(|2z^2| - 1) > |z|(2 - 1) = |z|, $$ contradicting the assumption. So all the roots have absolute value $\leq 1$. But the product of the absolute values of roots is $1$ (the constant term is $1$), so all of them are of absolute value one.

Again using the previous inequality with $|z| = 1$ instead of $|z| > 1$, we get that for all the roots $$ |2z^2 + 1| = |2z^2| - 1, $$ which is quite easily only possible if $z^2 = -1$ so $z = \pm i$. So only possible roots are $\pm i$ and if $i$ is, so is $2i^3 + i = -i$ and vice versa.

Lastly, to ensure that multiplicities of $i$ and $-i$ are equal: if $f$ has roots, it has factor $z^2 + 1$ by our observation, and factoring that out we see that the resulting polynomial also satisfies the functional equation. So indeed, only solutions are constant polynomials $0$ and $1$ and $(x^2+1)^k$ for positive integer $k$.

Another partial result: either $f(x) = 0 $ for all $x$ or $f$ has no real roots.

Proof:

Let $r$ be the largest real root of $f$. Then, by the functional equation, $ 2r^3+r$ is also a root of $f$, and so by the definition of $r$, $$ r \geq 2r^3+r $$ which simplifies to $$ 0 \geq r$$ Similarly, if we let $n$ be the smallest root of $f$, once again by the functional equation $2n^3+n $ is also a root and so $$ n \leq 2n^3+n$$ which simplifies to $$ 0 \leq n$$ So any root of $f$ must be between $0$ and $0$ and hence can only be $0$. So now there are two possibilities: 1) $f$ has no real roots 2) $f$ only has roots at $x=0$. In the second case, $f$ must be of the form $$ f(x) = a x^n $$ for some constant $a$ and non-negative integer $n$. Substituting this in the functional equation yields $$ ax^n a(2x^2)^n = a (2x^3+x)^n$$ $$ a^2 2^n x^{3n} = a (2x^3+x)^n $$ which is only possible if $n=0$ since otherwise the RHS has multiple terms. Hence, we are left with $$ a^2 = a$$ and $$f(x)=a$$ so $a=0$ or $a=1$, but we must take $a=0$ since case 2) is under the assumption that $f$ has a root at $x=0$. This completes the proof.