TITLE: Parameterize unitary without transpose
QUESTION [7 upvotes]: For all unitary matrices, i.e. $A \overline{A}^T = I$, there is a skew-Hermitian matrix $X$ so that $A = exp(X)$. So the unitary group has $n^2$ dimensions.
Is there any similar parameterisation of all matrices with $A \overline{A} = I$? What can be said about the number of dimensions? (Btw. is there any name for this type of matrices in literature?)

REPLY [3 votes]: I totally lied about this not being a natural thing to ask! As loup blanc alludes to, in fact $n \times n$ matrices such that $M^{-1} = \overline{M}$ can be interpreted as Galois descent data for descending the complex vector space $\mathbb{C}^n$ to a real vector space, or equivalently as a $1$-cocycle in 
$$Z^1(B \text{Gal}(\mathbb{C}/\mathbb{R}), GL_n(\mathbb{C})).$$
The statement that any such matrix must have the form $U \overline{U}^{-1}$ says that any such $1$-cocycle is cohomologous to zero, or equivalently that the Galois cohomology set
$$H^1(B \text{Gal}(\mathbb{C}/\mathbb{R}), GL_n(\mathbb{C}))$$
is trivial. This just says that there is only one real form of $\mathbb{C}^n$ up to isomorphism, namely $\mathbb{R}^n$. The space $GL_n(\mathbb{C})/GL_n(\mathbb{R})$ appearing in Robert Bryant's answer can then be interpreted as the space of real forms of $\mathbb{C}^n$.