User blog comment:Alemagno12/Some more set theory questions/@comment-1605058-20180104200708

Q4 fixed: Define a function \(f\) mapping the \(\alpha\)-th regular ordinal to \(\alpha\), where (mainly for convenience) we take \(\omega\) to be the first. I claim that, for any limit ordinal \(\alpha\), \(F(\aleph_{\alpha+n})=\omega f(\mathrm{cof}(\alpha)\)+n\). Hopefully it's clear we only need to show that for \(n=0\), as the remaining case follows from the successor case of the definition. I'll say it's clear for \(\alpha=\omega\), so assume it's larger

If \(\alpha\) is not a regular ordinal, then \(\beta=mathrm{cof}(\alpha)<\alpha\) and \(\mathrm{cof}(\aleph_beta)=\mathrm{cof}(\beta)=\beta=\mathrm{cof}(\alpha)=\mathrm{cof}(\aleph_\alpha)\), so we have \(F(\aleph_\alpha)=F(\aleph_\beta)=f(\mathrm{cof}(\beta))=f(\mathrm{cof}(\alpha))\).

If \(\alpha\) is a regular ordinal, then \(\aleph_\alpha\) has cofinality larger than any smaller cardinal, so \(F(\aleph_\alpha)\) is the least ordinal greater than \(F(\aleph_{\gamma+n})\) for \(\gamma<\alpha\) limit, so it's the least ordinal above \(\omega f(\mathrm{cof}(\gamma))+n\), which clearly is \(\omega f(\alpha)=\omega f(\mathrm{cof}(\alpha))\).

From this it's easy that the fixed points of \(F\) are (again!) weakly inaccessible cardinals.