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'.
Best Answer
Certain important properties are shared by all free ultrafilters. In many applications of ultrafilters, especially more elementary applications, only these properties are used. In such a situation, it does not matter which free ultrafilter is chosen -- any one will do.
But there are some proofs that require (or seem to require) special kinds of ultrafilters.
One important example, mentioned in the comments by Benjamin Steinberg, are algebraically special ultrafilters, such as the idempotent ultrafilters used to prove Hindman's Theorem, or the minimal idempotents used in the ultrafilters proof of the Hales-Jewett Theorem. These ultrafilters are "algebraically special" in the sense that they have special properties defined using the algebraic-and-topological structure $(\beta \mathbb N,+)$. However, if we don't want to look at all this extra structure on $\beta \mathbb N$, an idempotent ultrafilter may not be too different from any other. In fact, every idempotent is equivalent to many non-idempotents (in the sense of being isomorphic to them, as described in your first paragraph).
Another special kind of ultrafilter is a $P$-point. These are definable in the topological space $\beta \mathbb N$, without considering any other dynamic or algebraic structure, and a $P$-point cannot be isomorphic to a non-$P$-point. One important characterization of $P$-points is: an ultrafilter $\mathcal U$ is a $P$-point if and only if for any sequence $\langle x_n \rangle$ of real numbers, $r = \mathcal U$-$\lim x_n$ if and only if there is a subsequence of $\langle x_{n_k} \rangle$ converging to $r$ with $\{n_k :\, k \in \mathbb N\} \in \mathcal U$. I have seen this property of $P$-points used in proofs before, although no particularly famous examples spring to mind. I remember a theorem of mine, in a paper with Piotr Oprocha, where we need to take $\mathcal U$-limits that do not have this property. So this is an application of ultrafilters where it is important to use a non-$P$-point.
From an order-theoretic point of view, Ramsey ultrafilters are quite special. One can write a short proof of Ramsey's Theorem (the infinitary version) using a Ramsey ultrafilter. (I like this proof for pedagogical purposes, and have given it to graduate students in the past, since it nicely parallels the proof that measurable cardinals are Ramsey.)
So some applications of ultrafilters really do use special ultrafilters. To address some of your other questions:
It is a theorem of ZFC, not just a consistency result, that there are RK-incomparable ultrafilters. This is due to Kunen and Frolik.
It is also a theorem of ZFC that the action of homeomorphisms on $\beta \mathbb N \setminus \mathbb N$ is not transitive. In fact, Kunen proved (from ZFC only) that $\beta \mathbb N \setminus \mathbb N$ contains weak $P$-points. These are defined as points of $\beta \mathbb N \setminus \mathbb N$ that are not contained in the closure of any countable subset of $\beta \mathbb N \setminus \mathbb N$ not already containing the point. It isn't too hard to see that, because $\beta \mathbb N \setminus \mathbb N$ is compact, not every point has this property. And no self-homeomorphism of $\beta \mathbb N \setminus \mathbb N$ can map a weak $P$-point onto a non-weak $P$-point.
Finally, is there a meaningful sense in which all free ultrafilters are the same? Maybe. It is an open question whether all free ultrafilters can (consistently) have the same Tukey type. Roughly, there is a means of comparing partial orders via maps called Tukey reductions, much coarser than the notion of isomorphism described in your post. It is course enough that -- maybe -- all ultrafilters compare to all others. But this is consistently not the case. For example, a $P$-point is never Tukey-equivalent to a non-$P$-point.