$Tr(A^2)=Tr(A^3)=Tr(A^4)$ then find $Tr(A)$

A slightly different approach: we have $\operatorname{trace}((A-A^2)^2)=0$. As $A$ has a real spectrum, so does $A-A^2$. So, the previous trace condition implies that $A-A^2$ is nilpotent. Hence every eigenvalue of $A$ is equal to its square. As $A$ is nonsingular, every eigenvalue of $A$ must be equal to $1$. Hence the trace of $A$ is $n$.

Let's denote the eigenvalues by $t_j$. Then, by Cauchy-Schwarz $$ \sum t_j^3 \le \left( \sum t_j^2 \sum t_j^4\right)^{1/2} , $$ which equals $\sum t_j^3$ by assumption. Equality in the CSI means that the vectors are linearly dependent, so $t_j=ct_j^2$. This says that there is only one eigenvalue ($=1/c$), and clearly it must be $1$ then. So $\textrm{tr}\, A=n$.