Consider a category $\bf {Set}$ of sets and functions that admits a functor $^{*}-:\bf{Set}\to \bf {Set}$ which sends every set $S$ to an enlargement of it and every function $f:S\to T$ to its enlargement.

Is it possible that such an enlargement functor is essentially idempotent?

The problem is related to the existence of a special ultrafilter on $\mathbb {N}$. Let $g:\mathbb{N}\times \mathbb{N}\to \mathbb{N}$ be a bijection. If we write $UF(X)$ for the set of ultrafilters on the set $X$ then $g$ induces a bijection $G:UF(\mathbb{N}\times \mathbb{N})\to UF(\mathbb{N})$. Define the function $H_g:UF(\mathbb{N})\to UF(\mathbb{N})$ by sending an ultrafilter $\mathcal{F}\in UF(\mathbb{N})$ to $G(\mathcal{F}\times\mathcal{F})$. Call the pair $(g,\mathcal{F})$ 'good' if $H_g(\mathcal{F})=\mathcal{F}$ and $\mathcal{F}$ is non-principal.

Is there a good pair $(g,\mathcal{F})$?

Remark: The existence of a good pair $(g,\mathcal{F})$ is related to the enlargement problem as follows. Assume $ZF$ for the sets in $\bf {Set}$ and consider the usual ultrapower construction with respect to $\mathcal{F}$ to obtain an enlargement functor $^{*}-:\bf{Set}\to \bf {Set}$. $g$ can then rather straightforwardly be used to obtain the components of a natural isomorphism from the double enlargement to the enlargement.