Consider the set of ultrafilters $\beta(\mathbb N)$ on $\mathbb N$. Any function $f\colon\mathbb N\to\mathbb N$ extends to a function $\beta f\colon \beta \mathbb N \to \beta\mathbb N$. We say that two ultrafilters $\mathcal U$ and $\mathcal V$ are isomorphic if there is some bijection $f$ with $f(\mathcal U) = f(\mathcal V)$. Since there are only $2^{\aleph_0}$ many bijections of $\mathbb N$, but $2^{2^{\aleph_0}}$ many ultrafilters on $\mathbb N$, we know that there are many isomorphism classes of free ultrafilters.

On the other hand, in any proof that I have seen using ultrafilters, it does not seem to matter which ultrafilter is chosen. This leads me to the following Question: is there some way in which all free ultrafilters are the 'same'?

I have thought of some possibilities what it could mean for ultrafilters to be the 'same'. We can see any ultrafilter $\mathcal U$ as an ordered set, using the partial order $\subseteq$. I can imagine that if $\mathcal U$ and $\mathcal V$ are free ultrafilters, they are isomorphic as partial orderings. This seems pretty weak though.

Another possibility would be to consider the action of $\operatorname{Homeo}(\beta\mathbb N)$ on $\beta\mathbb N$. Does it act transitively?

It might be interesting to consider the Rudin–Keisler ordering $\leq_{\text{RK}}$ on $\beta\mathbb N$. It is defined by $\mathcal U\leq_{\text{RK}} \mathcal V$ iff there is a function $f\colon\mathbb N\to\mathbb N$ with $\beta f(\mathcal V) = \mathcal U$. It is known that there exist free ultrafilters that are not minimal for the Rudin–Keisler ordering, while it is independent of ZFC whether there exists free ultrafilters that are not minimal. Presumably, a minimal ultrafilter is not the 'same' as a not-minimal ultrafilter. However, even then it might be consistent with ZFC that all free ultrafilters are the 'same'.

Consider the set of ultrafilters $\beta(\mathbb N)$ on $\mathbb N$. Any function $f\colon\mathbb N\to\mathbb N$ extends to a function $\beta f\colon \beta \mathbb N \to \beta\mathbb N$. We say that two ultrafilters $\mathcal U$ and $\mathcal V$ are isomorphic if there is some bijection $f$ with $(\beta f)(\mathcal U) = \mathcal V$. Since there are only $2^{\aleph_0}$ many bijections of $\mathbb N$, but $2^{2^{\aleph_0}}$ many ultrafilters on $\mathbb N$, we know that there are many isomorphism classes of free ultrafilters.

On the other hand, in any proof that I have seen using ultrafilters, it does not seem to matter which ultrafilter is chosen. This leads me to the following Question: is there some way in which all free ultrafilters are the 'same'?

I have thought of some possibilities what it could mean for ultrafilters to be the 'same'. We can see any ultrafilter $\mathcal U$ as an ordered set, using the partial order $\subseteq$. I can imagine that if $\mathcal U$ and $\mathcal V$ are free ultrafilters, they are isomorphic as partial orderings. This seems pretty weak though.

Another possibility would be to consider the action of $\operatorname{Homeo}(\beta\mathbb N)$ on $\beta\mathbb N$. Does it act transitively?

It might be interesting to consider the Rudin–Keisler ordering $\leq_{\text{RK}}$ on $\beta\mathbb N$. It is defined by $\mathcal U\leq_{\text{RK}} \mathcal V$ iff there is a function $f\colon\mathbb N\to\mathbb N$ with $\beta f(\mathcal V) = \mathcal U$. It is known that there exist free ultrafilters that are not minimal for the Rudin–Keisler ordering, while it is independent of ZFC whether there exists free ultrafilters that are not minimal. Presumably, a minimal ultrafilter is not the 'same' as a not-minimal ultrafilter. However, even then it might be consistent with ZFC that all free ultrafilters are the 'same'.

Consider the set of ultrafilters $\beta(\mathbb N)$ on $\mathbb N$. Any function $f\colon\mathbb N\to\mathbb N$ extends to a function $\beta f\colon \beta \mathbb N \to \beta\mathbb N$. We say that two ultrafilters $\mathcal U$ and $\mathcal V$ are isomorphic if there is some bijection $f$ with $f(\mathcal U) = f(\mathcal V)$. Since there are only $2^{\aleph_0}$ many bijections of $\mathbb N$, but $2^{2^{\aleph_0}}$ many ultrafilters on $\mathbb N$, we know that there are many isomorphism classes of free ultrafilters.

On the other hand, in any proof that I have seen using ultrafilters, it does not seem to matter which ultrafilter is chosen. This leads me to the following Question: is there some way in which all free ultrafilters are the 'same'?

I have thought of some possibilities what it could mean for ultrafilters to be the 'same'. We can see any ultrafilter $\mathcal U$ as an ordered set, using the partial order $\subseteq$. I can imagine that if $\mathcal U$ and $\mathcal V$ are free ultrafilters, they are isomorphic as partial orderings. This seems pretty weak though.

Another possibility would be to consider the action of $\text{Homeo}(\beta\mathbb N)$ on $\beta\mathbb N$. Does it act transitively?

It might be interesting to consider the Rudin–Keisler ordering $\leq_{\text{RK}}$ on $\beta\mathbb N$. It is defined by $\mathcal U\leq_{\text{RK}} \mathcal V$ iff there is a function $f\colon\mathbb N\to\mathbb N$ with $\beta f(\mathcal V) = \mathcal U$. It is known that there exist free ultrafilters that are not minimal of the Rudin-Keisler ordering, while it is independent of ZFC whether there exists free ultrafilters that are not minimal. Presumably, a minimal ultrafilter is not the 'same' as a not-minimal ultrafilter. However, even then it might be consistent with ZFC that all free ultrafilters are the 'same'.

Consider the set of ultrafilters $\beta(\mathbb N)$ on $\mathbb N$. Any function $f\colon\mathbb N\to\mathbb N$ extends to a function $\beta f\colon \beta \mathbb N \to \beta\mathbb N$. We say that two ultrafilters $\mathcal U$ and $\mathcal V$ are isomorphic if there is some bijection $f$ with $f(\mathcal U) = f(\mathcal V)$. Since there are only $2^{\aleph_0}$ many bijections of $\mathbb N$, but $2^{2^{\aleph_0}}$ many ultrafilters on $\mathbb N$, we know that there are many isomorphism classes of free ultrafilters.

On the other hand, in any proof that I have seen using ultrafilters, it does not seem to matter which ultrafilter is chosen. This leads me to the following Question: is there some way in which all free ultrafilters are the 'same'?

I have thought of some possibilities what it could mean for ultrafilters to be the 'same'. We can see any ultrafilter $\mathcal U$ as an ordered set, using the partial order $\subseteq$. I can imagine that if $\mathcal U$ and $\mathcal V$ are free ultrafilters, they are isomorphic as partial orderings. This seems pretty weak though.

Another possibility would be to consider the action of $\operatorname{Homeo}(\beta\mathbb N)$ on $\beta\mathbb N$. Does it act transitively?

It might be interesting to consider the Rudin–Keisler ordering $\leq_{\text{RK}}$ on $\beta\mathbb N$. It is defined by $\mathcal U\leq_{\text{RK}} \mathcal V$ iff there is a function $f\colon\mathbb N\to\mathbb N$ with $\beta f(\mathcal V) = \mathcal U$. It is known that there exist free ultrafilters that are not minimal for the Rudin–Keisler ordering, while it is independent of ZFC whether there exists free ultrafilters that are not minimal. Presumably, a minimal ultrafilter is not the 'same' as a not-minimal ultrafilter. However, even then it might be consistent with ZFC that all free ultrafilters are the 'same'.