The least relation which produces a partial order Fix some set $U$. Let $E$ be a binary relation on $U$.
By definition $S(E)=\operatorname{id}_U \cup E \cup E^2 \cup E^3 \cup\dots$.
I define $\mu(E) = \bigcap \{X\in\mathscr{P}(U\times U) \mid S(X)=E \}$. That, informally speaking, is $\mu(E)$ is the least relation $X$ which produces $E$.
Question: Is $S(\mu(E)) = E$ for every partial order $E$?
If the question is false, does it become true if we consider only these $X$ which are partial orders themselves?
 A: The answer is no: it can happen that $S(\mu(E))\neq E$.
$\newcommand{\X}{\mathrel{X}}\newcommand{\E}{\mathrel{E}}$
As a counterexample, consider the usual ordering on $\mathbb{Q}$ (in fact any dense linear order will do). I claim that $\mu(\leq)=\emptyset$. To see this, let $a< b$ be rational numbers and $X=\,\leq\!\setminus\{(a,b)\}$. We get $S(X)=\,\leq$, since $a\X c\X b$ for any rational $a<c<b$, which puts $(a,b)$ into $S(X)$. So $X$ is one of the relations considered when computing $\mu(\leq)$, which means that $(a,b)\notin\mu(\leq)$.
As $\mu(\leq)=\emptyset$, it is now clear that $S(\mu(\leq))$ is the identity relation and, in particular, not equal to $\leq$.

Update: This is a fun problem. Here is a characterization of when your conclusion is true for linear orders.
A piece of notation, if $E$ is a partial order, I use $a\uparrow=\{b;a\E b\}$ for the cone above $a$ and $a\downarrow= \{b;b\E a\}$ for the cone below $a$. I also write $E^*$ for the reverse order.
Theorem: Let $E$ be a linear order. The following are equivalent:


*

*$S(\mu(E))=E$;

*$E$ does not admit suborders of type $\omega+1$ or $(\omega+1)^*$;

*For any $a$, the cones $a\uparrow$ and $a\downarrow$ are well-ordered by $E$ and $E^*$, respectively;

*$E$ embeds into $\mathbb{Z}$.


Proof:
($1\implies 2$): Suppose that $E$ has a suborder of type $\omega+1$ (the other case is similar). Enumerate this suborder as $a_0\E a_1\E \dots\E a_\omega$. For any $n$ let $E_n=E\setminus\{(a_n,a_\omega)\}$ and observe that $S(E_n)=E$. It follows that $(a_n,a_\omega)\notin \mu(E)$ for any $n$ and therefore $(a_n,a_\omega)\notin S(\mu(E))$ for any $n$.
($2\implies 3$): Clear.
($3\implies 4$): Assuming $E$ is nonempty, pick any $a$ and map it to $0$. Since the cones $a\uparrow$ and $a\downarrow$ are well-ordered, $a$ has an immediate $E$-successor and $E$-predecessor. Map these to $1$ and $-1$, respectively. Continue inductively. Note that we never run out of room, since any element we could not find an image for would have to have one of its two cones illfounded.
($4\implies 1$): For any $a$ let $a^-$ and $a^+$ be its immediate predecessor and successor, respectively. Observe that if $X$ is any relation such that $S(X)=E$ then $(a^-,a),(a,a^+)\in X$ for any $a$. Thus all these pairs are in $\mu(E)$ and it follows (since any two points of $\mathbb{Z}$ have finite distance) that $S(\mu(E))=E$. $\square$
The situation with partial orders seems to be less clear. For example, there are illfounded partial orders $E$ such that $S(\mu(E))=E$; take for example a copy of $(\omega+1)^*$ and on the side insert a single point between each $n$ and $\omega$.
