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This question was firstly posted on Mathoverflow. Two answers are pretty interesting. The potential counter-example given in the second answer is really interesting, but it is not surely a counter-example.

Consider two partially ordered sets $A = \{a< b,a< c\}$, $B=\{x< z,y< z\}$.

Their linear extensions (here we allow equality in linear extensions) for $A, B$ are $A_L=\{A_1=\{a< b< c\}, A_2=\{a< b= c\}, A_3=\{a< c< b\}\}$ and $B_L = \{ B_1 = \{x< y< z\}, B_2=\{y< x< z\}, B_3 =\{x= y< z\}\}$

We may define $f_1: A_1\to B_1$ by $f_1(a)=x, f_1(b)=y, f_1(c)=z$ and $f_2:A_3\to B_2$ by $f_2(a)=y, f_2(b)=x, f_2(c)=z$, but there are no bijective mappings from $A_2$ to $B_3$ s.t. $t< s\Longleftrightarrow f_3(t)< f_3(s)$ and $t=s \Longleftrightarrow f_3(t)=f_3(s)$. (And it's easy to see that no other pairing of the $A_i$ with the $B_j$ will allow such maps to be chosen.)

There are no order isomorphisms from $A$ to $B$, either.

Is the following conjecture true?

For any partially ordered sets $A,B$, if there exist isomorphisms from $A$'s linear extensions to $B$'s, then there exists an isomorphism from $A$ to $B$.

If not, could you give me a counter example?

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