User blog comment:Deedlit11/Ordinal Notations IV: Up to a weakly inaccessible cardinal/@comment-24509095-20140507175906/@comment-5150073-20140508111343

King, yes, it must be $$\Omega_\alpha$$. By the way, it applies always as I considered. $$\psi_I(\psi_I(\psi_I(\cdots ))) = \psi_I(\Omega_I) = \psi_I(I)$$ (note that I is the fixed point of $$f(\alpha) = \Omega_\alpha$$.