enlargements of sets and ultrafilters on countable sets

Question

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.

Answer 1

If one unwraps the packaging, it appears that the question about good pairs is asking the following:

Question. Is there a nonprincipal ultrafilter $F$ on $\mathbb{N}$ such that $F\times F$ is isomorphic to $F$ via a bijection $g:\mathbb{N}\times\mathbb{N}\to \mathbb{N}$?

The answer to this is no. I think about it in terms of seed theory: the ultrapower by $F\times F$ amounts to doing the ultrapower by $F$ twice, and therefore has two distinct seeds generating $F$, but the ultrapower by $F$ has a unique seed generating $F$, by the unique seed lemma, and so they cannot be isomorphic.

But let me also give a direct argument. Assume that $g:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ is an isomorphism of $F\times F$ with $F$. Let $p:\mathbb{N}\times\mathbb{N}\to \mathbb{N}$ be the projection function $p(x,y)=x$, and let $f=p\circ g^{-1}:\mathbb{N}\to\mathbb{N}$. Using the definition of the product ultrafilter, the fact that $g$ is an isomorphism, and the definition of $f$, it follows that $$A\in F\iff p^{-1}A\in F\times F\\ \iff g[p^{-1}A]\in F\\ \iff f^{-1}A\in F$$ Now, it is a completely general fact about ultrafilters, proved by Solovay, that if $A\in F\iff f^{-1}A\in F$ for all $A$, then $f$ must be the identity on a set in $F$. (I can post a proof of this if it is desired. This fact is the essence of the unique seed lemma.) Thus, $f(n)=n$ on a set $A$ in $F$. What this means is that $g^{-1}(n)$ is a pair with first coordinate $n$, for all $n\in A$. It follows that $g^{-1}A$ has at most one element on each column, and consequently is not in $F\times F$. Since $A\in F$, this contradicts the assumption that $g$ is an isomorphism of $F\times F$ with $F$. QED