User blog comment:Deedlit11/Ordinal Notations V: Up to a weakly Mahlo cardinal/@comment-30004975-20171217042147/@comment-1605058-20171217130704

If you define \(I_\alpha\) to be the \(\alpha\)-th inaccessible, then every fixed point of \(\alpha\mapsto I_\alpha\) must be inaccessible itself (since it's equal to \(I_\alpha\) after all!) so in particular it's regular.