Orthogonal and symmetric Matrices What can one say about the set of all $n$-dimensional square matrices $A \in \text{GL}_n(\mathbb{C})$ that have an inverse with entries out of $\mathbb{C}$ with the properties:


*

*unitary $:\Leftrightarrow A^*= A^{-1}$

*hermitian $:\Leftrightarrow A^* = A$


where $A^*$ is the conjugate transpose of $A$.
What obviously follows is $$A^{-1} = A$$
The most simple matrix that is in this set is the identity matrix. Are there others? How do they look like?
 A: (in addition to Mariano's answer), these are sometimes called Householder transformations. 
A: A simple answer for $A \in \mathbb{R}^{2 \times 2}$:
Yes, there are more matrices. They are of the form
$A = \begin{pmatrix}\cos \alpha & \sin \alpha\\
\sin \alpha & -\cos \alpha \end{pmatrix}$.
It is obvious that these matrices are symmetric.
The inverse is $\frac{1}{(\cos \alpha) \cdot (-\cos \alpha) - sin^2 \alpha} \begin{pmatrix}-\cos \alpha & -\sin \alpha\\
-\sin \alpha & \cos \alpha \end{pmatrix} = \frac{-1}{\cos^2 \alpha + sin^2 \alpha} \begin{pmatrix}-\cos \alpha & -\sin \alpha\\
-\sin \alpha & \cos \alpha \end{pmatrix} = A$
Maybe you can find such forms for $\mathbb{C}^{n \times n}$, too.
I know this is not exactly an answer to your question. But I think this explicit, simple form is a nice answer for a part of the question.
A: Such a matrix is such that $A^2=I$, so it is diagonalizable, and its possible eigenvalues are $+1$ and $-1$. Since it is unitary, the eigenspaces corresponding to $1$ and to $-1$ are orthogonal. 
Conversely, every diagonalizable matrix with eigenvalues contained in $\{+1,-1\}$ and orthogonal eigenspaces is of that form.
It follows that the set of your matrices is in bijection with the set of subspaces of $\mathbb C^n$. Explicitely: If $V$ is one such subspace, there is a unique linear transformation $f:\mathbb C^n\to\mathbb C^n$ such that $V$ and $V^\perp$ are eigenspaces for the eigenvalues $1$ and $-1$. The matrix $A_V$ of $f$ with respect to the standard basis of $\mathbb C^n$ satisfies your conditions. The bijection is $$V\in\{\mathrm{subspaces\;of\;}\mathbb C^n\}\leftrightarrow A_V.$$
As a consequence, considered as a whole, your set is a disjoint union of submanifolds homeomorphic to Grassmannian varieties. Literally books have been written about them.
