(1) I am interesting in finding a rank 3 unitary matrix $U$ satisfying $$ U U^*=-1=-\text{diag}(1,1,1). $$ $$ U U^\dagger=+1=+\text{diag}(1,1,1). $$
(2) I am interesting in finding a rank 4 unitary matrix $U$ satisfying $$ U U^*=-1=-\text{diag}(1,1,1,1). $$ $$ U U^\dagger=+1=+\text{diag}(1,1,1,1). $$
Here $*$ for complex conjugate, and the $\dagger$ for complex conjugate then transpose.
This question is secondary, or not necessary if you can answer (1) and (2):
(3) I wonder what are the generic properties of such $U$? Can it be embed into some continuous Lie group? [part of $U(N)$?]
Let $n$ be the rank of $U$. For $n=3$ (or $n$ odd in general) there is no solution as the determinant of $UU^*$ cannot be negative. For $n$ even there are many solutions. For example, for $n=4$, the following $U$, with $\theta$ and $\phi$ arbitrary angles, satisfies the requirements of (2):
$$U=\begin{pmatrix} 0&-e^{i\theta}&0&0\\ e^{i\theta}&0&0&0\\ 0&0&0&-e^{i\phi}\\ 0&0&e^{i\phi}&0\\ \end{pmatrix}\,, $$
What you want is a $U$ with the following property: if $v$ is an eigenvector of $U$ with eigenvalue $e^{i\chi}$, then $v^*$ should be also and eigenvector of $U$ with eigenvalue $-e^{i\chi}$.