I was going to prove the following statement:
Let $F$ be a normal function, then $\cup\{0,F(0),F(F(0)),\ldots\}$ is the least ordinal $\alpha$ satisfying $F(\alpha)=\alpha$.
In case of $0\neq F(0)$ it is certainly true that none of $F(0),F(F(0)),\ldots$ can be the least fixed point so the above statement would be satisfied. But what if $0=F(0)$? What if $F$ in fact is the identity? Then we had $\cup\{0,F(0),F(F(0)),\ldots\}=\{0\}$ which would not be the least fixed point. /edit: no, see comments and answer below …
So the identity function is a counterexample, or is the identity not a normal function for some reason I am missing?!
Yes the identity is a normal function, and $\cup\{0\}=\{0\}$ is just rubbish as Noble Mushtak pointed out in the comments.
To prove the cited statement, tick the case $F(0)=0$ as trivial. If $F(0)\neq 0$, suppose $\cup\{0, F(0), F(F(0)),\ldots\}$ was not the least fixed point. Then $F^i(0)$ for some $i\geq 1$ is the least fixed point but then $F^{i-1}(0)<F^i(0)$ whence $F^i(0)<F^{i+1}(0)$ as $F$ is normal. Contradiction!