Does every Partially Ordered relation and its dual have the same number of topological orderings?
Is it the case that both the poset and its dual poset have the same number of topological orderings?


Yes, you have observed/discovered a fact about partially ordered set (poset). In fact, there is a stronger and more pleasing fact, which is a special case of the duality of order theory.



The dual order of a topological ordering of a poset is a topological order of the dual order of that poset.

Given a poset $p$, let $p^{d}$ be the dual order of $p$.  Let $p\rightarrowtail$q mean $q$ is a topological ordering of $p$. Then the above claim can be expressed as 

$$ p\rightarrowtail q\ \Leftrightarrow p^d \rightarrowtail q^d$$

Proof: All we need to do is to follow the definitions. Let the underlying set of $p$ be $S$, which is also the underlying set of $q$, $p^d$ and $q^d$. For a poset $p$, we use "$\preceq_p$" to denote the binary relation on of $p$ $S\times S$.

"$\Rightarrow$":


Let $s_1,s_2\in S$ such that $s_1\preceq_{p^d} s_2$. Then by the definition of the dual order, $s_2\preceq_{p} s_1$. By the definition of a topological sort, $s_2\preceq_{q} s_1$. By the definition of the dual order, $s_1\preceq_{q^d}s_2$. 
Since $q$ is a total order, so is $q^d$.


"$\Leftarrow$": 


Let $s_1,s_2\in S$ such that $s_1\preceq_{p} s_2$. Then by the definition of the dual order, $s_2\preceq_{p^d} s_1$. By the definition of a topological sort, $s_2\preceq_{q^d} s_1$. By the definition of the dual order, $s_1\preceq_{q}s_2$. 
Since $q^d$ is a total order, so is ${(q^d)}^d=q$.


In fact, the implication "$\Leftarrow$" is the dual of the implication "$\Rightarrow$". Replacing $p$ by $p^d$ and $q$ by $q^d$ in the implication "$\Rightarrow$", we get "$\Leftarrow$". 

Q.E.D.



The claim above shows that given a poset $p$, whenever we can construct a topological order $q$ for $p$, we also have $q^d$, a topological order for $q^d$.
Hence $q\to q^d$ maps the topological orders of $p$ to the topological orders of $p^d$.

Similarly, whenever we can construct a topological order $r$ for $p^d$, we also have $r^d$, a topological order for ${(p^d)}^d=p$.

The map $q\to q^d$ and $r\to r^d$ are inverse functions to each other. So, each map is, in fact, a one-one correspondence. So the number of topological orderings of $p$ equals the number of the topological orderings of $p^d$.



Be careful that the topological orders of the original poset $p$ and its dual $p^d$ might not be same, although they are dual to each other. 

For example, we can check the poset $p=\{a,b,c\}$ with $a\preceq_p b$ and $a\preceq_p c$. All topological orders of $p$ shares the unique least element $a$ but don't share greatest elements. All topological orders of $p^d$ shares the unique greatest element $a$ but don't share least elements.



Exercise. A poset is connected if the directed graph representing the poset is connected. Show that the maximum number of topological orderings of a connected poset of $n$ vertices is $(n-1)!$.