Let matrix $A$ be normal. If $A{A^T}$ has $n$ distinct eigenvalues, why is $A$ symmetric? Let matrix $A \in {M_n}(\Bbb R)$ be normal. If $A{A^T}$ has $n$ distinct eigenvalues, why is $A$ symmetric?
Here's a late answer.

First, a bit of notation. Since we will work over $\mathbb C,$ we will need the notion of "conjugate-transpose" of a complex matrix. Namely, if $M\in \mathbb C^{n\times n},$ then its "conjugate-transpose" is defined as
$$
M^H = (\overline M)^T,
$$
where $\overline \cdot$ denotes complex conjugation of the matrix entries and $\cdot^T$ denotes matrix transposition. This is a well known operation. Observe that $\overline \cdot$ and $\cdot^T$ commute. And, of course, for $N\in\mathbb R^{n\times n},$ we have $N^H = N^T.$ In the following, we will use only the operation $\cdot^H.$

On to the actual question. So we are given a matrix $A\in\mathbb R^{n\times n}$ which is normal, i.e. $A^HA = AA^H$ and such that $A^HA$ has $n$ distinct eigenvalues. We want to show that $A$ is symmetric, i.e. $A^H = A.$

Observe that, since $A$ is real, its characteristic polynomial has real coefficients. This means that the eigenvalues of $A,$ i.e. the roots of its characteristic polynomial, are either real or come in complex conjugate pairs. In other words, we can enumerate the eigenvalues of $A$ as a sequence
$$
\lambda_1,\ldots,\lambda_r,\mu_1,\overline{\mu_1},\ldots,\mu_s,\overline{\mu_s},
$$
with $r,s\in \mathbb N_0,$ $\lambda_j\in\mathbb R,$ and $\mu_k\in \mathbb C$ with $\Im\mu_k &gt; 0.$ Repetitions are allowed, i.e. we can have $\lambda_j = \lambda_k$ or $\mu_j = \mu_k$ for $j\neq k.$

Now, since $A$ is normal, it can be diagonalized using a unitary transformation. This is a well known fact. In more detail, we have the following. Put
$$
D = {\rm diag}(\lambda_1,\ldots,\lambda_r,\mu_1,\overline{\mu_1},\ldots,\mu_s,\overline{\mu_s})\in \mathbb C^{n\times n},
$$
i.e. $D$ is the diagonal matrix with the eigenvalues of $A$ on the diagonal. Then, there is a unitary matrix $U\in \mathbb C^{n\times n}$ such that
$$
\tag{1}
A = U^HDU.
$$
Recall that "unitary" means
$$
\tag{2}
U^HU = UU^H = I,
$$
where $I \in \mathbb C^{n\times n}$ is the identity matrix. From $(1),$ we get
$$
\tag{3}
A^H = (U^HDU)^H = U^HD^HU.
$$
From $(1),$ $(2),$ and $(3),$ we conclude
$$
A^HA = \left(U^HD^HU\right)\left(U^HDU\right) = U^HD^HDU.
$$
This shows that the eigenvalues of $A^HA$ are the diagonal elements of the dialgonal matrix
$$
\begin{align}
D^HD &amp; = {\rm diag}(\lambda_1^2,\ldots,\lambda_r^2,\overline{\mu_1}\mu_1,\mu_1\overline{\mu_1},\ldots,\overline{\mu_s}\mu_s,\mu_s\overline{\mu_s}) \\
&amp; = {\rm diag}(\lambda_1^2,\ldots,\lambda_r^2,|\mu_1|^2,|\mu_1|^2,\ldots,|\mu_s|^2,|\mu_s|^2).
\end{align}
$$
Observe that every pair $\mu_j,\overline{\mu_j}$ of complex conjugate eigenvalues of $A$ produces twice the same eigenvalue for $A^HA.$ By assumption, this can't happen. So we must have that $s = 0,$ $A$ has only real eigenvalues, and in particular $D\in \mathbb R^{n\times n},$ i.e. $D$ is real. But then
$$
\tag{4}
D^H = D.
$$
Now we get from $(1),$ $(3),$ and $(4)$
$$
A^H = U^HD^HU = U^HDU = A,
$$
as desired.
$A$ is normal and $AA^T$ has $n$ distinct eigenvalues, so $AA^T=X \Lambda^2X^{-1}=A^TA $, where $\Lambda$ is diagonal and has $\lambda$'s on the diagonal.

Now $AA^T=(X \Lambda X^{-1})(X \Lambda X^{-1})=A^TA $ so that $A=A^T$.