Why is there an Zariski open set of $P\in GL(n,\mathbb{R})$ such that $P\,\text{diag}(1,\dots,1,-1)P^{-1}$ can be conjugated by a diagonal matrix $D$ to get an orthogonal matrix?
Note that $M=P\,\text{diag}(1,\dots,1,-1)P^{-1}$ is involutory ($MM=I$) and $\det(M)=-1$.
Possible starting point: Consider $$\{P\in GL_n(\mathbb R),\ D \text{ invertible and diagonal s.t. } DP\text{diag}(1,\dots,1,-1)P^{-1}D^{-1} \text{ is orthogonal}\}$$
In general, this gives $n^2$ polynomial equations of order 2 in $n+n^2$ variables (the coefficients of $P$ and $D$). We can go a bit further by noticing that the condition of orthogonality can be reduced to $\frac{n(n-1)}{2}$ conditions. I have the feeling that using the fact that the eigenspace $E_{-1}$ of $M$ is of dimension $1$ and $E_1$ is of dimension $n-1$ would reduce the number of "required" polynomial equations down to $n$.
Then, as the invertible matrices are Zariski dense in $M_n(\mathbb R)$ and so $GL_n(\mathbb R)$ has dimension $n^2$ (as a group), we could conclude that the remaining $n$ degrees of freedom can be used to choose such a $D$. But I am not sure...
