To be reflexive, it must include all pairs $(a,a)$ with $a\in A$. To be symmetric, whenever it includes a pair $(a,b)$, it must include the pair $(b,a)$. So it amounts to choosing which $2$-element subsets from $A$ will correspond to associated pairs. If you pick a subset $\{a,b\}$ with two elements, it corresponds to adding both $(a,b)$ and $(b,a)$ to your relation.
How many $2$-element subsets does $A$ have? Since $A$ has $n$ elements, it has exactly $\binom{n}{2}$ subsets of size $2$.
So now you want to pick a collection of subsets of $2$-elements. There are $\binom{n}{2}$ of them, and you can either pick or not pick each of them. So you have $2^{\binom{n}{2}}$ ways of picking the pairs of distinct elements that will be related.
Your assumption that the number of (symmetric and reflexive) relations equals the number of (purely reflexive) relations is wrong.
Let me explain:
Say, $A=\{1,2\}$
Reflexive relations on $A$ are
$\{(1,1),(2,2)\}$,
$\{(1,1),(2,2),(1,2)\}$,
$\{(1,1),(2,2),(2,1)\}$,
$\{(1,1),(2,2),(1,2),(2,1)\}$
Thus the number of reflexive relations equals 4 ($2^{n(n-1)}$ in general).
But the number of reflexive and symmetric relations equals $2^{\frac{n(n-1)}{2}}$ as is already described in the link you've provided.
The number of reflexive relations is always greater than the number of reflexive and symmetric relations.
And in your example, it's not just the principle diagonal. You've neglected the symmetric pairs that can exist along with the ordered pairs necessary to make the relation a reflexive one, i.e $\{(1,1),(2,2),(3,3),(4,4),(5,5),(1,2),(2,1)\}$ is also both reflexive and symmetric.
Best Answer
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.