Matrices which are both unitary and Hermitian

Unitary matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are on the unit circle. Hermitian matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are real. So unitary Hermitian matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are $\pm 1$.

This is a very strong condition. As George Lowther says, any such matrix $M$ has the property that $P = \frac{M+1}{2}$ admits a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are $0, 1$; thus $P$ is a Hermitian idempotent, or as George Lowther says an orthogonal projection. Of course such matrices are interesting and appear naturally in mathematics, but it seems to me that in general it's more natural to start from the idempotence condition.

I suppose one could say that Hermitian unitary matrices precisely describe unitary representations of the cyclic group $C_2$, but from this perspective the fact that such matrices happen to be Hermitian is an accident coming from the fact that $2$ is too small.

A matrix $M$ is unitary and Hermitian if and only if $M=2P-1$ for an orthogonal projection $P$. That is, $P$ is Hermitian and $P^2=P$.

Since no-one else seems to have said it (explicitly at least, although elements of order $2$ and projections are closely linked, as indicated in some answers), a unitary matrix which is also Hermitian is just a unitary matrix of multiplicative order at most $2$ (or, equivalently, a Hermitian matrix of multiplicative order at most $2$). For a matrix $A$ is unitary if an only if $A^{*} = A^{-1},$ where $*$ denotes "transposed conjugate", while $A$ is Hermitian if and only if $A^{*} = A.$ Hence if $A$ is both unitary and Hermitian, we have $A = A^{-1}$ (and $A$ is unitary). As for theoretical uses, the group ${\rm SU}_{n}^{\pm}(\mathbb{C})$ is generated by such matrices for every $n$, where ${\rm SU}_{n}^{\pm}(\mathbb{C})$ denotes the group of unitary $n \times n$ matrices of determinant $\pm 1$. This is clear for $n = 1$, and follows easily by induction, using the fact that ${\rm PSU}(n,\mathbb{C})$ is a simple group for $n > 1.$