Why the paired orbit has the same size here? enter link description here
On this proof, he just showed that the the paired orbit of (a,b) has the same size, but this seem has nothing to do with the size of corresponding paired orbit of an orbit of Gs on S. 
In addition, how to show when |G| is odd, then we don't have an element of G interchanges a and b?  
 A: I posted an answer in the link (to your previous question) by mistake.  I'll copy it here:
Let $G$ act on a set $S$.  The orbits of the induced action of $G$ on $S \times S$ are called the orbitals of $G$.  Fix $s \in S$. It can be shown that if the action of $G$ on $S$ is transitive, then there is a 1-1 correspondence between the orbits of the action of $G_s$ on $S$ and the orbitals of $G$.  The orbital $\Delta:=(s,t)^G$ corresponds to the $G_s$-orbit $\{t: (s,t) \in \Delta \}$.  One way to visualize this is using the tabular representation of relations on $S$: each orbital corresponds to a union of permutation matrices, and the columns corresponding to the nonzero coordinates in row $s$ of an orbital form the corresponding $G_s$-orbit.  
An orbital $(s,t)^G$ is said to be self-paired if it equals its transpose $(t,s)^G$.  If this is the case, then some element of $G$ interchanges $s$ and $t$. This element contains the 2-cycle permutation $(s,t)$ and hence has even order.  Thus $G$ also has even order.  Conversely, if $G$ has even order, then by Cauchy's theorem it has an element of order 2, which contains a 2-cycle $(x,y)$. Thus, there exists an orbital $(x,y)^G$ that equals its transpose.  We have shown that $G$ has even order if and only if some orbital is self-paired.  Take the contrapositive, and we get that $G$ has odd order if and only if every orbital is not equal to its transpose.  In particular, for each orbital $\Delta$ of $G$, there is another orbital $\Delta^*$ which is disjoint from $\Delta$ but which has the same order as $\Delta$.  Due to the 1-1 correspondence, these two orbitals give rise to two distinct $G_s$-orbits of the same order.  For example, if $\Delta$ is a union of $k$ permutation matrices, then so is $\Delta^*$, and the two corresponding $G_s$-orbits have $k$ elements each.
