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. 



enlargements of sets and ultrafilters on countable sets
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


