On normal matrices and norms Let $A^*$ denote the complex conjugate transpose of the matrix $A$ and $\|\cdot\|=\|\cdot\|_2$ the norm induced by the Euclidean vector norm.
If
$$
\|A^*A+AA^*\|-\|A^*A-AA^*\|=\|A^*A\|
$$
what can be said about $A$? Note that $A$ does not have to be normal, in fact if $A$ is normal and satisfies the above condition, then $A=0$, as noted in the comments below.
My first hypothesis, after lots of numerical experiments was that $\det(A)=0$, since almost all random matrices with the above property fulfilled had zero determinant, but then I found
$$
A=\begin{pmatrix}-2&2&0\\0&0&4\\-2&-2&0\end{pmatrix}
$$
for which $\det(A)=-32$. This shattered my hypothesis, but the question remains. What can be said about such matrices?
EDIT: I'm putting a small bounty on this question. Let's define a class $\mathcal{A}$ of complex-valued matrices such that
$$
\mathcal{A}=\{A\in\mathbb{C}^{n\times n}\;:\; \|A^*A+AA^*\|-\|A^*A-AA^*\|=\|A^*A\|\}\;.
$$
Can you find any necessary or sufficient conditions for a matrix to be in this class? Special cases, $A\in\mathbb{R}^{n\times n}$ for a fixed $n$ are welcome.
 A: Note first that $A\in {\mathcal A}$ if and only if $\lambda A\in {\mathcal A}$ for any non-zero number $\lambda$. Hence
there is no loss of generality if we are concerned with matrices $A$ which have norm $1$. Since $\mathbb{C}^{n\times n}$
is a $C^*$-algebra if it is endowed with the operator norm, one has $\| A^* A\|=\| A\|^2$. Hence we are looking for
matrices $A\in \mathbb{C}^{n\times n}$ such that 
$$ \| A\|=1\quad  \text{and}\quad \|A^*A+AA^*\|-\|A^*A-AA^*\|=1. $$
First we are concerned with  matrices of rank $1$. We will use the notation from operator theory: $A=u\otimes v$, where $u, v \in {\mathbb C}^n$ 
are nonzero vectors. The action of $A$ on ${\mathbb C}^n$ is given by 
$$Ax=\langle x,v\rangle u\qquad (x\in {\mathbb C}^n), $$
where $\langle \cdot, \cdot \rangle$ is the inner product in ${\mathbb C}^n$
(linear in the first component and antilinear in the second).
Since it is assumed that $\| A\|=1$ we can assume also that $\| u\|=\| v\|=1$. If $u$ and $v$ were linearly dependent, then
$A$ would be normal and hence it would not be in ${\mathcal A}$. Hence we assume that $u$ and $v$ are linearly independent vectors of norm $1$. 
We denote $\omega=\langle u,v\rangle$. Observe that $|\omega|< 1$ (since $u$ and $v$ are linearly independent of norm $1$).
Note that $A^*=v\otimes u$, $AA^*=u\otimes u$ and $A^*A=v\otimes v$. Of course, $AA^*+A^*A=u\otimes u+v\otimes v$ is a positive semidefinite
matrix of rank $2$. Hence its norm is equal to the largest eigenvalue. Since the range of $AA^*+A^*A$ is the linear span of vectors $u$ and $v$
the eigenvectors corresponding to non-zero eigenvalues of $AA^*+A^*A$ are of the form $x=\alpha u+\beta v$. Let $\lambda>0$ be an eigenvalue of
$AA^*+A^*A$ and $x=\alpha u+\beta v$ be an eigenvector at this eigenvalue. Then
$$[u\otimes u+v\otimes v](\alpha u+\beta v)=\lambda (\alpha u+\beta v)$$
gives
$$ (\alpha +\beta \overline{\omega}-\lambda \alpha)u+ (\alpha \omega +\beta -\lambda \beta)v=0. $$
Since $u, v$ are linearly independent we have
$$ \alpha +\beta \overline{\omega}-\lambda \alpha=0\quad \text{and}\quad \alpha \omega +\beta -\lambda \beta=0. $$
We have $\alpha \ne 0$ or $\beta \ne 0$ and therefore one can deduce that $\lambda=1\pm |\omega|$. It follows that
$$ \|A^*A+AA^*\|=1+|\omega|. $$
Matrix $A^*A-AA^*=u\otimes u-v\otimes v$ is selfadjoint and of rank $2$. In a similar way as before we see that the nonzero eigenvalues
of this matrix are $\pm \sqrt{1-|\omega|^2}$. Hence $\|A^*A-AA^*\|=\sqrt{1-|\omega|^2}$. 
Finally, it follows from
$$ 1=\|A^*A+AA^*\|-\|A^*A-AA^*\|=1+|\omega|-\sqrt{1-|\omega|^2}$$
that
$$ |\omega|=\frac{\sqrt{2}}{2}. $$
It is not hard to check now that any $A=u\otimes v$ with $\| u\|=1=\| v\|$ and $|\langle u,v\rangle|=\frac{\sqrt{2}}{2}$ belongs to ${\mathcal A}$.
Let now $A\in \mathbb{C}^{n\times n}$ be a matrix of rank $1$ in ${\mathcal A}$, i.e., $A=u\otimes v$ with $\| u\|=1=\| v\|$ and $|\langle u,v\rangle|=\frac{\sqrt{2}}{2}$,
and let $B\in \mathbb{C}^{m\times m}$ be any matrix such that 
$$\| B\| \leq 1,\quad \| BB^*+B^*B\| \leq 1+\frac{\sqrt{2}}{2}\quad \text{and}\quad \| BB^*-B^*B\| \leq \frac{\sqrt{2}}{2}. $$
Then, for $A\oplus B\in \mathbb{C}^{(n+m)\times (n+m)}$, one has 
$$\| A\oplus B\| =1,\quad \| (A\oplus B)(A\oplus B)^*+(A\oplus B)^*(A\oplus B)\| = 1+\frac{\sqrt{2}}{2}$$
and
$$ \| (A\oplus B)(A\oplus B)^*-(A\oplus B)^*(A\oplus B)\| = \frac{\sqrt{2}}{2}, $$
which means that $A\oplus B$ belongs to ${\mathcal A}$ (in dimension $n+m$).
