If ${\scr U}$ and ${\scr V}$ are ultrafilters on non-empty sets $A$ and $B$ respectively, then the *tensor product* ${\scr U}\otimes{\scr V}$ is the following ultrafilter on $A\times B$:
$$\big\{X\subseteq A\times B: \{a\in A:\{b\in B: (a,b)\in X\}\in {\scr V}\}\in {\scr U}\big\}.$$
Is there a non-principal ultrafilter ${\scr U}$ on $\omega$ and a function $f:\omega \to (\omega\times\omega)$ such that $f({\scr U}) = {\scr U}\otimes {\scr U}$?



Relation between ultrafilters ${\scr U}$ and ${\scr U} \otimes {\scr U}$
As Will Brian noted in a comment, either projection $p:\omega^2\to\omega$ satisfies $p(\mathcal U\otimes\mathcal U)=\mathcal U$. If you also had a function $f$ as in the question, with $f(\mathcal U)=\mathcal U\otimes\mathcal U$, then the composite function $f\circ p$ would send $\mathcal U\otimes\mathcal U$ to itself. By a result that I quoted at [Selective ultrafilter and bijective mapping](https://mathoverflow.net/questions/324254), and for which Ali Enayat supplied a reference, it would follow that $f\circ p$ becomes the identity function when restricted to a suitable set in $\mathcal U\otimes\mathcal U$. That is absurd, because $p$ is not one-to-one on any set in $\mathcal U\otimes\mathcal U$.


