Ask 12 Let $f(x)=\sqrt{x^{2}+1}-1$. Find all real solutions to the equation $f(\underbrace{f(\cdots(f(x))}_{260} \cdots$
1)) $=x$.
(1) The solution to $f(x)=x$ must also be a solution to $f^{[200]}(x)=x$, so $M_{1} \subseteq M_{200}$.
(2) Assume $\exists x_{0} \in M_{200}$, but $x_{0} \notin M_{1}$. For $\forall x_{1}, x_{2}$, we have
$$
f\left(x_{1}\right)-f\left(x_{2}\right)=\sqrt{x_{1}^{2}+1}-\sqrt{x_{2}^{2}+1}=\frac{\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)}{\sqrt{x_{1}^{2}+1}+\sqrt{x_{2}^{2}+1}} \text {. }
$$

Therefore, when $x_{1}>x_{2} \geqslant 0$, $f\left(x_{1}\right)>f\left(x_{2}\right)$, meaning $f(x)$ is increasing on the interval $[0,+\infty]$.
Also, when $x>0$, $f(x)-x=\sqrt{x^{2}+1}-(x+1)f(x) > 0$, thus, when $x>0$,
$$
x>f(x)>f(f(x))>\cdots>f^{[200]}(x)>0 .
$$

And when $x<0$, similarly, $f(x)>f(f(x)) \cdots>f^{[200]}(x)>0$.
Thus, $M_{200}=M_{1}$, and $\sqrt{x^{2}+1}-1=x \Rightarrow x=0$, so $M_{200}=\{0\}$.