Suppose $f$ is a $\kappa$ -normal function and consider any $\alpha<\kappa$ and define a sequence by $\alpha_0=\alpha$ and $\alpha_{n+1}=f(\alpha_n)$ . Let $\alpha_\omega=\sup_{n<\omega}\alpha_n$ . Then, since $f$ is continuous, $$f(\alpha_\omega)=\sup_{n<\omega} f(\alpha_n)=\sup_{n<\omega}\alpha_{n+1}=\alpha_\omega$$ So $\operatorname{Fix}(f)$ is unbounded.

Suppose $N$ is a set of fixed points of $f$ with $|N|<\kappa$ . Then $$f(\sup N)=\sup_{\alpha\in N} f(\alpha)=\sup_{\alpha\in N}\alpha=\sup N$$ so $\sup N$ is also a fixed point of $f$ , and therefore $\operatorname{Fix}(f)$ is closed.