User blog comment:Alemagno12/Question/@comment-35470197-20180730031159

On Attempt 2:

I note that you are not allowed to use in the usual set theory.
 * 1) "a class \(F\) of functions \(textrm{Ord} \to \textrm{Ord}\)"
 * 2) the condition "if there is some subclass of"

For 1, every class consists of sets, and every function \(textrm{Ord} \to \textrm{Ord}\) is a class which is not a set. Therefore there is not such a class.

For 2, the quantification (\(\exists\)) of classes is not allowed.

But in this case, you can safely state a legal formula with the essentially same meaning. So I do not mind it here.

Well, anyway, the answer is no. The emptyset satisfies the second condition, and hence \(FF = \emptyset\).

Moreover, a function does not necessarily admit a fixed point. Say, \(x \mapsto x+1\) does not. Maybe you need to replace the second condition in an appropiate way. At least, the answer is still no, even if you assume the first condition on \(FF\).