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.
EDIT: The original question has been changed, so my answer refers to the question "is the relation 'has the same parents as' symmetric, reflexive, or transitive?"
Let A, B, and C be people.
For part a):
Symmetric: If A has the same parent as B, then does B has the same parents as A? Yes, so it is symmetric.
Reflexive: Does A have the same parents as A? Obviously yes, so it's reflexive.
Transitive: If A has the same parents as B, and B has the same parents as C, then does A have the same parents as C? Yes, so it is transitive.
Can you figure out b) and c)?
Best Answer
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.