Is every finite connected set equipped with a binary relation of the form Aut(X)\X? Is every finite connected set equipped with a binary relation of the form Aut(X)\X?
Say that a related set is a couple $(X,R_X)$ where $X$ is a set and $R_X\subset X^2$ a binary relation on $X$. Abusing the notation we often write $X$ instead of $(X,R_X)$ and $R$ instead of $R_X$. We also abbreviate the condition $(x,y)\in R$ by $xRy$. If $Y$ is another related set we say that a map $f:X\to Y$ is a morphism if $xRx'$ implies $f(x)Rf(x')$ for all $x,x'\in X$. We define the notions of isomorphism and automorphism in the usual way. We say that $X$ is connected if the equivalence relation generated by $R$ has exactly one class.
Let $X$ be a finite connected related set, $G$ its automorphism group and $X_*$ the set of all $G$-orbits in $X$. For $A,B\in X_*$ decree that $ARB$ holds in $X_*$ if and only if $aRb$ holds in $X$ for some $a\in A,b\in B$. Equipped with this relation $X_*$ is again a finite connected related set, whence the question

Is every finite connected related set $X$ isomorphic to $Y_*$ for some finite connected related set $Y$?

[Without the connectedness assumption, the related set with two elements and no relation would be a counterexample.]
Clearly, if $X$ is a poset, so is $X_*$.
This answer shows that for every connected poset $X$ there is a connected poset $Y$ such that $Y_*\simeq X$.
Proposition. Every finite connected related set $X$ of size at most 3 is isomorphic to $Y_*$ for some finite connected related set $Y$.
Let us introduce some notation for the statement of Lemma 1 below.
Assume that $m,n\in\mathbb Z_{\ge0}$ with $m\ge1$; that $a_1$, ... $a_m$, $b_1$, ..., $b_m$, $c_1$, ..., $c_n$, $b'_1$, ..., $b'_m$, $c'_1$, ..., $c'_n$ are distinct elements; that the set $X$ defined by
$$
X=\{a_1,\ldots,a_m,b_1,\ldots,b_m,c_1,\ldots,c_n\};
$$
is equipped with a binary relation; and that the $\operatorname{Aut}(X)$-orbits in $X$ are the $\{a_i,b_i\}$ and the $\{c_j\}$. Form the related set
$$
X'=\{a_1,\ldots,a_m,b'_1,\ldots,b'_m,c'_1,\ldots,c'_n\}
$$
such that the obvious bijection $X\to X'$ is an isomorphism. Equip the set $Y:=X\cup X'$ with the unique binary relation such that
(a) the binary relations of $X$ and $X'$ are induced by that of $Y$,
(b) if $x\in X\setminus X'$ and $x'\in X'\setminus X$ then $x$ and $x'$ are unrelated in $Y$.
The proof of Lemma 1 and Lemma 2 below a tedious but easy exercise using the list of isomorphism classes of related sets of size 3 given on this page, which is part of the entry number of nonisomorphic unlabeled binary relations on n labeled nodes of The On-Line Encyclopedia of Integer Sequences.
Lemma 1. In the above setting, assume that $X$ is connected and that $2m+n\le3$. Then $Y$ is a finite connected related set satisfying $Y_*\simeq X$.
Lemma 2. Let $X=\{a_1,a_2,a_3\}$ be a related set of size three and $G$ its automorphism group. Assume that $G$ acts transitively on $X$. Let $Y=\{a_{11},a_{21},a_{22},a_{31},a_{32},a_{33}\}$ be a set of size $6$. Define the binary relation $R$ on $Y$ by
$$
a_{ij}Ra_{pq}\iff
\begin{cases}
a_iRa_p&\text{if }i\ne p\\ 
j=q\text{ and }a_iRa_i&\text{if }i=p.
\end{cases}
$$
Then $Y_*$ is isomorphic to $X$.
I haven't been able to decide if Lemma 2 holds without the assumption that $2m+n\le3$.
The proposition follows from the lemmas.
 A: This is a partial answer, too long for a comment.
If $X$ is a poset we write $X^-$ for the subset of minimal elements, and $X^+$ for the subset of maximal elements. An element of $X^-\cap X^+$ is called an isolated point.
Proposition. If $X$ is a nonempty finite poset without isolated points, then there is a finite poset $Y$ without isolated points such that $Y_*\simeq X$.
To prove the proposition we will define a poset $Y$ and show that it has the required properties.
Denote by $|S|$ the cardinality of any set $S$ and put $[n]:=\{1,2,\ldots,n\}$ for any nonnegative integer $n$. Let $X$ be a nonempty finite poset without isolated points. Set
$$
m:=|X^-|,\quad n:=|X|,\quad X^-=\{x_1,\ldots,x_m\}.
$$
Let $x_{m+1},x_{m+2},\ldots,x_n$ be an enumeration of the elements of $X\setminus X^-$ such that $x_i<x_j$ implies $i<j$.
We will first define integers $c_1<\cdots<c_n<r$ and then define $Y$ using these integers.
For $i\in[m]$ we set $c_i:=i$.
$\bullet$ Definition of $c_{m+1}<c_{m+2}<\cdots<c_n$: Let $\Pi$ be the set of all non-constant polynomial $P(t)\in\mathbb Q[t]$ with positive leading coefficient; here $t$ is an indeterminate. Set for $m<i\le n$
$$
P_i(t):=\sum_{x_j<x_i}\binom tj,
$$
where the sum runs over the $j\in[m]$ such that $x_j<x_i$. Since $P_i\in\Pi$ for all $i$, there are integers $c_{m+1},\ldots,c_n\in m\mathbb Z$ such that $m<c_{m+1}<\cdots<c_n$ and $P_i(c_i)\ne P_j(c_j)$ whenever $m<i<j\le n$.
$\bullet$ Definition of $r$: For $i\in[m]$ denote by $\Pi_i$ the set of all the polynomials in $\Pi$ whose degree is congruent to $-i$ modulo $m$, and set
$$
Q_i(t):=\sum_{x_j>x_i}\binom{t-i}{c_j-i},
$$
where the sum runs over the $j\in[n]$ such that $x_j>x_i$. Since each $Q_i$ is in $\Pi_i$, the $Q_i$ are pairwise distinct. Hence there is an integer $r>c_n$ such that the $Q_i(r)$ are pairwise distinct.
$\bullet$ Definition of $Y$: For $i\in[n]$ set
$$
Y_i:=\{S\subset[r]\ |\ |S|=c_i\},\quad Y:=\bigcup Y_i.
$$
For $S\in Y_i$, $T\in Y_j$ set
$$
S<T\iff x_i<x_j\text{ and }S\subset T.
$$
Then $\le$ is a partial order on $Y$. Note that for $S\in Y_i$ we have: $S$ is minimal in $Y$ if and only if $x_i$ is minimal in $X$ and $S$ is maximal in $Y$ if and only if $x_i$ is maximal in $X$. In particular $Y$ is a nonempty finite poset without isolated points.
$\bullet$ Proof of the isomorphism $Y_*\simeq X$: Write $S\sim T$ for $S,T\in Y$ to indicate that there is an automorphism of $Y$ which maps $S$ to $T$.
There is a strictly increasing surjection $f:Y\to X$ mapping $S\in Y_i$ to $x_i$. We claim that $f$ induces an isomorphism $Y_*\to X$. To prove this it suffices to show that for $S\in Y_i$ and $T\in Y_j$ we have $S\sim T$ if and only if $i=j$. If $i=j$, then, since $S$ and $T$ are two cardinality $c_i$ subsets of $[r]$, there is a permutation $\sigma$ of $[r]$ which moves $S$ to $T$, and $\sigma$ induces an automorphism of $Y$ which maps $S$ to $T$. It only remains to prove that $S\sim T$ implies $i=j$.
Recall that we have $S\in Y_i$, $T\in Y_j$, $S\sim T$ and we claim $i=j$.
Case 1: $S$ is minimal. Then $T$ is also minimal and we get $i,j\in[m]$. The number of elements of $Y$ which are greater than $S$ (respectively greater than $T$) is $Q_i(r)$ (respectively $Q_j(r)$). But the assumption $S\sim T$ implies $Q_i(r)=Q_j(r)$, and thus $i=j$.
Case 2: $S$ is not minimal. Then $T$ is also not minimal and we get $m<i,j\le n$. The number of minimal elements of $Y$ which are less than $S$ (respectively less than $T$) is $P_i(c_i)$ (respectively $P_j(c_j)$). But the assumption $S\sim T$ implies $P_i(c_i)=P_j(c_j)$, and thus $i=j$. This completes the proof of the proposition.
