$f\left(x\right)$ has extreme points at $\{-1, 0, 1\}$

$\Rightarrow f^{\text{'}}\left(x\right)=0$ at $x= –1, 0, 1$

$\Rightarrow f^{\text{'}}\left(x\right)=a\left(x^2-1\right)x \because f \left(x\right)$ is a polynomial

$\Rightarrow f\left(x\right)=\int a\left(x^3-x\right)dx=a\left(\frac{x^4}{4}-\frac{x^2}{2}\right)+ C$

Now the equation,

$f\left(x\right)=f\left(0\right)$

$\Rightarrow a\left(\frac{x^4}{4}-\frac{x^2}{2}\right)+ C=C$
$\Rightarrow \frac{x^4}{4}-\frac{x^2}{2}=0 \Rightarrow x^2\left(x^2-2\right)=0$
$\Rightarrow x=0, \pm \sqrt{2}$

$\Rightarrow S= \{0, \sqrt{2}, -\sqrt{2}\}.$

Hence, the set contains two irrational and one rational number.