Let $X=\left(\begin{array}{ll}a& b\\ c& d\end{array}\right) \& A={\left(\begin{array}{ll}0& i\\ 1& 0\end{array}\right)}^n$

$\Rightarrow AX=IX$

$\Rightarrow A=I$

$\Rightarrow {\left(\begin{array}{ll}0& i\\ 1& 0\end{array}\right)}^n=I$

$\Rightarrow A^8=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow n$ is multiple of $8.$

So number of $2$ digit numbers in the set 

$S=\left\{16,24,32,\dots ,96\right\}$

Clearly, $16,24,32,\dots ,96$ are in arithmetic progression.

Here, $a=16, d=24-16=8$ and $l=a+\left(n-1\right)d=96$

Where, $a$ is the first term, $d$ is the common difference, $l$ is the last term and $n$ is the number of terms.

Consider, $l=a+\left(n-1\right)d=96$

$\Rightarrow 16+\left(n-1\right)8=96$

$\Rightarrow \left(n-1\right)8=80$

$\Rightarrow \left(n-1\right)=10$

$\Rightarrow n=11$

Hence, number of two-digit numbers in set $S$ is $11$.