<s>[INST] How to prove that $A^2$ is a symmetric matrix? Conjecture 1 :

Let $A$ be a real matrix such that $A^5=A A^T A A^T A$.  Then $A^2$ is a symmetric matrix.

(here $A^T$ denotes the transpose of a matrix A).
I guess that the following is also true : 
Conjecture 2 :

If $A^{2n+1}=AA^TAA^T\cdots AA^TA$ then 
  $A^n $ is symmetric.

PS: This second conjecture has been shown to be false when $A$ is invertible, see Robert Israel's answer below. But I still think that it's true when $A$ is not invertible.
The post if matrix such $AA^T=A^2$ then $A$ is symmetric? solves the $n=1$ case.
 [/INST] Assume that $A\in M_3(\mathbb{R})$. Since the cases when $rank(A)=1,3$ are solved, assume that $rank(A)=2$. 
Prop. Under the hypothesis above, $A^5=AA^TAA^TA$ implies that $A^2$ is symmetric.
Proof. Let $A=\begin{pmatrix}a&b&c\\d&e&f\\g&h&k\end{pmatrix}$. We may assume that  $AA^T=diag(u,v,0)$ where $u,v>0$. Clearly $g=h=k=0$ and $A^2=\begin{pmatrix}a^2+bd&b(a+e)&ac+bf\\d(a+e)&bd+e^2&cd+ef\\0&0&0\end{pmatrix}$.
Since $A$ is defined up to a real factor, we may assume $u=1$ and finally we study the system $AA^T=diag(1,v,0)$, $A^5=\begin{pmatrix}a&b&c\\v^2d&v^2e&v^2f\\0&0&0\end{pmatrix}$. We use the Grobner basis software of Maple ; here the difficulty is that there are an infinity of solutions over $\mathbb{C}$ and over $\mathbb{R}$ and consequently, we must work also with hand !
We obtain $ad+be+cf=0$ and $d(v^2-1)=b(v^2-1)=cf(v^2-1)=0$.
Case 1. $v=1$. Then $a^2+b^2+c^2=d^2+e^2+f^2=1$ and $(c^2+f^2)(c^2+f^2-2)=0$.
Case 1.1. $c^2+f^2=2$. Then $c=\pm 1,f=\pm 1$, $a=b=d=e=0$ that is contradictory because $ad+be+cf\not=0$.
Case 1.2. $c^2+f^2=0$. Then $A^2$ is symmetric.
Case 2. $v\not=1$ and $d=b=cf=0$. We obtain $c^2(c^2-2)=0,a^2+c^2=1$ that implies $c=0,a^2=1$.
Case 2.1. $f=0$. then $A^2$ is symmetric.
Case 2.2. $f\not=0$. Then $2e^2+f^2=0$, that is contradictory.
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