Wacky Polynomial

Let $f(x) = x\left( 4x^2-3 \right) \left( 64x^6-96x^4+36x^2-3 \right)$. Find the last three digits of the number of distinct positive reals $x$ satisfying $f^{(2014)}(x) = x$.

Details and Assumptions

• $f^{(2014)}(x)$ denotes the function $f$ applied $2014$ times successively on $x$, i.e. $f^{(2014)}(x) = \underbrace{f ( f ( \cdots ( f(x) )\cdots ))}_{2014 \text{ times}}$.

• Any non-real / repeating solutions of the equation should be ignored. For example, the roots of the equation (counted with multiplicity) $x^5-14 x^4 + 86 x^3 - 298 x^2 + 573 x - 468=0$ are $\left \{ 3, 3, 4, 2-3i, 2+3i \right \}$, but its number of distinct real solutions is $2$.

This problem appeared in the Proofathon Algebra contest, and was posed by me.