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This is a great question. There has been quite a bit of work done to figure out what the continuous images of $\beta \mathbb N \setminus \mathbb N$ are. I'll do my best to summarize some of that work …

The answer is no, because every infinite closed subset of $\beta \omega$ has cardinality $2^\mathfrak{c}$. So for example, if $\{x_1,x_2,\dots\}$ is any countably infinite set of non-weak-$P$-points, …

Yes, this is possible. First of all, let me suggest a way of thinking about the $+$ operation. If you're familiar with the idea of taking limits along an ultrafilter, then given $p,q \in \mathbb Z^*$ …

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 …

An answer to question 2: Shelah constructed a model with exactly one $P$-point (up to isomorphism). In fact, the one $P$-point is a selective ultrafilter. You can find the construction in section XVII …

The answer to your main question is yes. In fact, there is (under $\mathsf{CH}$) a self-homeomorphism of $\omega^*$ with exactly one fixed point. Such a mapping is constructed in the proof of Theorem …

Yes, these two structures are elementarily equivalent. This is proved as a corollary to another theorem, which states Theorem: CH implies that $\Phi$ and $\Phi^{-1}$ are conjugate to each other in the …

Update: The answer is yes -- if $\mathsf{CH}$ is true then $\phi$ and $\phi^{-1}$ are conjugate in the group of self-homeomorphisms of $\omega^*$. I've written this up in a new paper, which you can fi …