Reducible finite order mappings classes and their action on the Thurston boundary

Question

Let $S$ denote the closed oriented surface of genus $g\geq 2$, and $\text{Mod}(S)$ be the mapping class group of $S$. Let $f\in \text{Mod}(S)$ be a finite order reducible element i.e. $f$ has a representative $\phi\in \text{Homeo}^+(S)$ such that $\phi^k=$ identity for some $k>1$ and $\phi$ preserve a multi-curve in $S$. There is a natural action of the cyclic group $\text{Mod}(S)$ on the Thurston boundary $\mathcal{PMF}(S)$. Is the set $$\text{Fix}(f):=\{[(\mathcal{F},\mu)]\in \mathcal{PMF}(S)|\, f.[(\mathcal{F},\mu)]=[(\mathcal{F},\mu)]\},$$ always a finite set?

Answer 1

No. If $f$ is the identity mapping class then $\mathrm{Fix}(f)$ is all of $\mathcal{PMF}(S)$.

More generally, if $f$ is periodic the quotient $X = S / f$ is an orbifold. Therefore we can take any projective measured foliation on $X$ and lift it back to $S$ to obtain a projective measured foliation on $S$ that is invariant under $f$. However, since $f$ is reducible, $X$ is not a triangle orbifold and so $X$ has infinitely many different projective measured foliations.