The answer is NO. Suppose that \operatorname{tr}(\mathrm{A})=0$and$A^{2}+A^{t}=I$. Taking the transpose, we have$$A=I-\left(A^{2}\right)^{t}=I-\left(A^{t}\right)^{2}=I-\left(I-A^{2}\right)^{2}=2 A^{2}-A^{4}$$$$A^{4}-2 A^{2}+A=0$$The roots of the polynomial$x^{4}-2 x^{2}+x=x(x-1)\left(x^{2}+x-1\right)$are$0,1, \frac{-1 \pm \sqrt{5}}{2}$so these numbers can be the eigenvalues of$A$; the eigenvalues of$A^{2}$can be$0,1, \frac{1 \pm \sqrt{5}}{2}$. By \operatorname{tr}(A)=0$, the sum of the eigenvalues is 0 , and by \operatorname{tr}\left(A^{2}\right)=\operatorname{tr}\left(I-A^{t}\right)=3$ the sum of squares of the eigenvalues is 3 . It is easy to check that this two conditions cannot be satisfied simultaneously.