First of all find $A\times A$ wherein $A=\{1,2,3\}$. So, we have $$A\times A=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}$$ Now try to choose some subset of above set as relations on $A$, since we know that every relation on $A$ has the form $R\subseteq A\times A$. For example $R_1=\{(1,1),(2,2),(3,3)\}$ is reflexive. $R_2=\{(1,1),(1,2),(2,1),(2,2)\}$ is reflexive, transitive and symmetric. Now try to find other relations according to what you have learnt about them.
We need to find the number of relations on $X$ that are reflexive, symmetric and anti-symmetric. Since the relation is reflexive, it contains the diagonal elements $\{(x,x): x \in X\}$. Since the relation is symmetric, if it contains an off-diagonal element $(x,y)$, where $x \ne y$, then it must also contain its transpose $(y,x)$. But since the relation is also antisymmetric, if it contains an off-diagonal element $(x,y)$, then it must NOT contain its transpose $(y,x)$. The only way both these conditions can be satisfied is that the relation not contain any off-diagonal elements at all in the first place. Hence the relation is exactly the set of diagonal elements. Thus there is only one relation on $X$ that is reflexive, symmetric and anti-symmetric.
Best Answer
I take the definition of an antisymmetric relation $R$ to mean that $a R b$ and $b R a$ implies $a = b$, but for a given $a$ and $b$ it might well be that neither $a R b$ nor $b R a$.
So the number should be $$ 2^{n} 3^{\binom{n}{2}}, $$ because you have two choices for pairs $(a, a)$ (either $a R a$ or not) while for pairs $(a, b)$, with $a < b$, you have three mutually exclusive choices, either $a R b$ or $b R a$ or neither.