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 5.7 in my paper:

"$P$-sets and minimal right ideals in $\mathbb N^*$," Fundamenta Mathematicae 229 (2015), pp. 277-293. (pdf)

Sketch of the construction: Begin with the shift map $\sigma$ on $\omega^*$, which is the (unique) self-homeomorphism of $\omega^*$ that is induced by the mapping $n \mapsto n+1$ on $\omega$. (That is, $\sigma$ maps every ultrafilter $\mathcal U$ to the ultrafilter generated by $\{ A+1 :\, A \in \mathcal U\}$.) This map has no fixed points. Using some ideas from earlier in the paper, there is a nowhere-dense, closed $P$-set $X \subseteq \omega^*$ such that $X$ is closed under $\sigma$. (Recall that $X$ is a $P$-set if the intersection of any countably many neighborhoods of $X$ is again a neighborhood of $X$.) Now modify the topology of $\omega^*$ by collapsing $X$ to a point. Assuming $\mathsf{CH}$, this quotient space is again homeomorphic to $\omega^*$. ($\mathsf{CH}$ implies that collapsing a nowhere dense $P$-set to a point always results in a copy of $\omega^*$; see Corollary 1.2.4 in van Mill's survey article about $\beta \omega$ for a proof.) The map $\sigma$ induces a self-homeomorphism of the quotient space, and the collapsed copy of $X$ is its unique fixed point. $\quad \square$

Notice that the construction is quite flexible. If you wanted exactly $17$ fixed points, this could be arranged. If you wanted no fixed points but a point of order $42$, then this could be arranged as well. (In the above argument, instead of collapsing $X$ to a point, collapse each of the $42$ sets $X \cap (k+42\mathbb N)^*$ to a point.) I think the most general thing one could achieve along these lines is: for any function $f: \mathbb N \rightarrow \mathbb N \cup \{\aleph_0\}$, there is (assuming $\mathsf{CH}$) a self-homeomorphism of $\omega^*$ having $f(n)$ points of order $n$ for every $n$.

For the bonus questions, I think this should be possible using a different technique (but I'd have to check the details to be sure). Under $\mathsf{CH}$ I think that any point $u \in \omega^*$ is a butterfly point, meaning that $\omega^* \setminus u$ can be partitioned into two nonempty clopen sets. And I think these two sets are homeomorphic, so permuting them (while leaving $u$ fixed) gives you the kind of map you want. Note that this also gives you a different way of getting a positive answer to your main question as well.

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 5.7 in my paper:

"$P$-sets and minimal right ideals in $\mathbb N^*$," Fundamenta Mathematicae 229 (2015), pp. 277-293. (pdf)

Sketch of the construction: Begin with the shift map $\sigma$ on $\omega^*$, which is the (unique) self-homeomorphism of $\omega^*$ that is induced by the mapping $n \mapsto n+1$ on $\omega$. (That is, $\sigma$ maps every ultrafilter $\mathcal U$ to the ultrafilter generated by $\{ A+1 :\, A \in \mathcal U\}$.) This map has no fixed points. Using some ideas from earlier in the paper, there is a nowhere-dense, closed $P$-set $X \subseteq \omega^*$ such that $X$ is closed under $\sigma$. (Recall that $X$ is a $P$-set if the intersection of any countably many neighborhoods of $X$ is again a neighborhood of $X$.) Now modify the topology of $\omega^*$ by collapsing $X$ to a point. Assuming $\mathsf{CH}$, this quotient space is again homeomorphic to $\omega^*$. ($\mathsf{CH}$ implies that collapsing a nowhere dense $P$-set to a point always results in a copy of $\omega^*$; see Corollary 1.2.4 in van Mill's survey article about $\beta \omega$ for a proof.) The map $\sigma$ induces a self-homeomorphism of the quotient space, and the collapsed copy of $X$ is its unique fixed point. $\quad \square$

Notice that the construction is quite flexible. If you wanted exactly $17$ fixed points, this could be arranged. If you wanted no fixed points but a point of order $42$, then this could be arranged as well. (In the above argument, instead of collapsing $X$ to a point, collapse each of the $42$ sets $X \cap (k+42\mathbb N)^*$ to a point.) I think the most general thing one could achieve along these lines is: for any function $f: \mathbb N \rightarrow \mathbb N \cup \{\aleph_0\}$, there is (assuming $\mathsf{CH}$) a self-homeomorphism of $\omega^*$ having $f(n)$ points of order $n$ for every $n$.

For the bonus questions, I think this should be possible using a different technique (but I'd have to check the details to be sure). Under $\mathsf{CH}$ I think that any point $u \in \omega^*$ is a bowtie point, meaning that $\omega^* \setminus u$ can be partitioned into two nonempty clopen sets. And I think these two sets are homeomorphic, so permuting them (while leaving $u$ fixed) gives you the kind of map you want. Note that this also gives you a different way of getting a positive answer to your main question as well.