jmc

If the leading term of $f(x)$ is $a{x}^m$, then the leading term of $f(x)f(2x^2)$ is $$a{x}^m\cdot a(2x^2{)}^m=2^m{a}^2{x}^{3m},$$and the leading term of $f(2x^3+x)$ is $2^{m}a{x}^{3m}$. Hence $2^m{a}^2=2^{m}a$, and $a=1$.

Because $f(0)=1$, the product of all the roots of $f(x)$ is $\pm 1$. If $f(\lambda )=0$, then $f(2{\lambda }^3+\lambda )=0$. Assume that there exists a root $\lambda$ with $|\lambda |\ne 1$. Then there must be such a root ${\lambda }_1$ with $|{\lambda }_1|>1$. Then $$|2{\lambda }^3+\lambda |\ge 2|\lambda {|}^3-|\lambda |>2|\lambda |-|\lambda |=|\lambda |.$$But then $f(x)$ would have infinitely many roots, given by ${\lambda }_{k+1}=2{\lambda }_k^3+{\lambda }_k$, for $k\ge 1$. Therefore $|\lambda |=1$ for all of the roots of the polynomial.

Thus $\lambda \overline{\lambda }=1$, and $(2{\lambda }^3+\lambda )\overline{(2{\lambda }^3+\lambda )}=1$. Solving these equations simultaneously for $\lambda =a+bi$ yields $a=0$, $b^2=1$, and so ${\lambda }^2=-1$. Because the polynomial has real coefficients, the polynomial must have the form $f(x)=(1+x^2{)}^n$ for some integer $n\ge 1$. The condition $f(2)+f(3)=125$ implies $n=2$, giving $f(5)=\boxed{676}$.