If the leading term of $ f(x) $ is $ a x^m $, then the leading term of $ f(x)f(2x^2) $ is \[ax^m \cdot a(2x^2)^m = 2^ma^2x^{3m},\]and the leading term of $ f(2x^3 + x) $ is $ 2^max^{3m} $. Hence $ 2^ma^2 = 2^ma $, 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 | \neq 1 $. Then there must be such a root $ \lambda_1 $ with $ |\lambda_1|>1 $. Then \[|2\lambda^3+\lambda | \geq 2|\lambda |^3-|\lambda | > 2|\lambda |-|\lambda |= |\lambda |.\]But then $ f(x) $ would have infinitely many roots, given by $ \lambda_{k+1}=2\lambda_k^3+\lambda_k $, for $ k \geq 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 \geq 1 $. The condition $ f(2) + f(3) = 125 $ implies $ n = 2 $, giving $ f(5) = \boxed{676} $.