My friend discovered a different approch of this kind of problem such as:
If A is a real matrix such that $(AA^T)^n=A^{2n}$ or $(AA^T)^nA=A^{2n+1}$, then $A^n=(A^T)^n.$
by making use of trace as follows:
For $AA^T=A^2,$ we can consider $\text{tr}((A-A^T)(A-A^T)^T)=\text{tr}(-A^2-(A^T)^2+A^TA+AA^T)=0 \implies A=A^T$
For $AA^TA=A^3$, we can consider the inequality $\text{tr}((AA^T-A^TA)^2)\geq 0 \implies \text{tr}(AA^TAA^T)\geq \text{tr}(A^2(A^T)^2).$ Note that$\text{tr}((A^2-AA^T)((A^T)^2-AA^T))=\text{tr}(A^2(A^T)^2)+\text{tr}{(AA^TAA^T)}-2\text{tr}(A^3A^T)\leq2\text{tr}(AA^TAA^T)-2\text{tr}(A^3A^T)=0,$ so we have $A^2=AA^T$ and this is what we have done.
For $AA^TAA^T=A^4,$ we can consider the inequality $\text{tr}(A^2-(A^T)^2)(A^2-(A^T)^2)\leq0 \implies \text{tr}(A^2(A^T)^2)\geq\text{tr}(A^4), $so actually we have a inequality chain $\text{tr}(AA^TAA^T)\geq\text{tr}(A^2(A^T)^2)\geq\text{tr}(A^4)$. However we have $AA^TAA^T=A^4,$ so the inequalities are equations. Hence $\text{tr}(A^2-(A^T)^2)(A^2-(A^T)^2)=0\implies A^2=(A^T)^2.$
...
We are interested about if these inequalities holds for every positive integer $n$:
If A is a real matrix, then $\text{tr}((AA^T)^n)\geq\text{tr}(A^{2n}). $
As what we proved above, for $n=1,2$, the statement is right. But we have no idea for $n\geq3$.
Can anyone help us to prove or disprove this inequality?
