The number $3$ plays an interesting role in the following statement:

$\newcommand{\S}{\sf(S_3)}\S$ Let $X$ be a non-empty set and let $f:X \to X$ be fixpoint-free (that is $f(x) \neq x$ for all $x\in X$). Then there are subsets $X_1, X_2, X_3 \subseteq X$ with $X_1\cup X_2\cup X_3 = X$ and $$X_i \cap f(X_i) = \emptyset$$ for $i \in \{1,2,3\}$.

There are easy examples showing that one cannot get by using $2$ subsets only. Statement $\S$ can be proved using the axiom of choice. (Although the full force of (AC) is not needed as the strictly weaker Boolean Prime Ideal Theorem (BPI) is sufficient, but it appears that the $\S$ is not a theorem of $\newcommand{\ZF}{{\sf (ZF)}}\ZF$.)

The motivation of this question is to create a theorem of $\ZF$ in the spirit of $\S$.

Question. Is the following statement a theorem of $\ZF$?

There is a cardinal $\kappa_0 \geq 3$ with the following property: If $X$ is a set with more than $1$ element and $f:X\to X$ is fixpoint-free, then there is a ${\frak S}\subseteq {\cal P}(X)$ with $\bigcup{\frak S} = X$, an injection $\iota:{\frak S}\to \kappa_0$, and the property that $S\cap f(S) = \emptyset$ for all $S\in {\frak S}$.

If the above statement is not a theorem of $\ZF$, then I am wondering about the following special case in $\newcommand{\N}{\mathbb{N}}\N$:

If $f:\N\to\N$ is fixpoint-free, then there is a finite set ${\frak S}\subseteq {\cal P}(\N)$ with $\bigcup{\frak S} = \N$ and $S\cap f(S) = \emptyset$ for all $S\in {\frak S}$.

(Note that if we drop the finiteness requirement on ${\frak S}$, then ${\frak S} = \big\{\{n\}:n\in\mathbb{N}\big\}$ is a trivial solution.)