It seems like maybe $u$ has to equal $v$, or maybe $ u = cv $ for some $ c > 0 $. Is this true?
If $ u v^* $ is positive semidefinite, then $ x^* u v^* x \ge 0$. I observe then that $ x^* u v^* x = (x^* u) (v^* x)$, the product of two inner products. If this is greater than equal to zero, then the two inner products must always have the same sign, for all $x$. Geometrically, this means $u$ and $v$ are always "pointing in the same direction" as $x$, for all $x$ (or both pointing "away" from $x$). I think this implies necessarily that $ u = cv $ for some $ c > 0 $. If not, we could choose an $x$ pointing in the same direction as $u$ but not $v$, and then $ (x^* u) (v^* x) < 0 $, violating the positive semidefinite condition.
Notice that the matrix annihilates the orthogonal complement of $v,$ and sends $v$ itself to a positive multiple of $u.$ It's adjoint annihilates everything orthogonal to $u$ and sends $u$ itself to a positive multiple of $v.$ So, the matrix is only self-adjoint when $u$ is a real multiple of $v,$ and only PSD if the multiple is positive. In which case it is just a positive multiple of the projection onto the subspace generated by $u$ (or $v$ if you prefer).