Consider the set of all permutations of $n$ numbers $\mathfrak{S}(n)$. Each permutation can be seen as a total ordering relation of $n$ elements $a_1,...,a_n$, such that \begin{equation} a_{\pi(1)}<a_{\pi(2)}<...<a_{\pi(n)}. \end{equation} Specific subsets of the set of permutations can be used instead to identify partial orderings. For example, for $n=3$, the three permutations $S=\{\pi_1,\pi_2,\pi_3\}\subset \mathfrak{S}(3)$ defined by \begin{align} \pi_1(1)=1, \ \pi_1(2)=2, \ \pi_1(3)=3, \\ \pi_2(1)=2, \ \pi_2(2)=1, \ \pi_2(3)=3, \\ \pi_3(1)=2, \ \pi_3(2)=3, \ \pi_3(3)=1, \end{align} identify the orderings \begin{align} a_1<a_2<a_3, \ a_2<a_1<a_3, \ a_2<a_3<a_1. \end{align} which are equivalent, when taken together, to the partial ordering \begin{equation} a_2<a_3. \end{equation} Another example is given by the two total orderings \begin{equation} a_1<a_2<a_3, \ a_2<a_1<a_3, \end{equation} which correspond to the partial ordering \begin{equation} a_1<a_3, \ a_2<a_3. \end{equation}
Any other subset of permutations, $S\subseteq \mathfrak{S}(n)$ can be certainly be divided into disjoint unions of sets $S_1,...,S_n$ such that $S_i$ corresponds to a partial order (the trivial decomposition is $S=\{\pi_1,...,\pi_m\}$ and $S_i=\{\pi_i\}$ $i=1,...,m$, in which case each sets corresponds to a total order).
My question is the following: is there a way to decompose $S$ into a minimal number of sets $S_1,...,S_n$ corresponding to partial orderings (minimal with respect to $n$), and is there a unique way to achieve this decomposition? According to this decomposition, the set of all permutations $\mathfrak{S}(n)$ corresponds to the partial ordering in which no element is actually in any ordered relation with any other element. If represented as a graph, it would have nodes but no links.