The number of $2\times 2$ matrices $A$, with each element as a real number and satisfying $A+A^T=I$ and $A^{T}A=I$, is:

Let the matrix $A$ be defined as:
$$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$$
Then its transpose $A^T$ is:
$$A^T=\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]$$
Given $A+A^T=I$:
$$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]+\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$
$$\left[\begin{array}{cc}2a& b+c\\ b+c& 2d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$
Equating corresponding elements:$2a=1⟹a=\frac{1}{2}$$b+c=0⟹b=-c$$2d=1⟹d=\frac{1}{2}$ Also given $A^{T}A=I$:
$$\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}{a}^2+c^2& ab+cd\\ ab+cd& b^2+d^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$
Substituting values of $a,d,b$:$a^2+c^2=1⟹(\frac{1}{2}{)}^2+c^2=1⟹\frac{1}{4}+c^2=1⟹c^2=\frac{3}{4}⟹c=\pm \frac{\sqrt{3}}{2}$ Since $b=-c$, $b=\mp \frac{\sqrt{3}}{2}$.The condition $ab+cd=0$ becomes $(\frac{1}{2})(-\frac{\sqrt{3}}{2})+(\frac{\sqrt{3}}{2})(\frac{1}{2})=0$, which is satisfied.The possible matrices are:
$$\left[\begin{array}{cc}1/2& \sqrt{3}/2\\ -\sqrt{3}/2& 1/2\end{array}\right]\text{ and }\left[\begin{array}{cc}1/2& -\sqrt{3}/2\\ \sqrt{3}/2& 1/2\end{array}\right]$$
Total number of matrices = $2$.