Let's say we have a point $nP = (x,y)$ on a curve $E$ over a prime $p$. The corresponding Edwards curve coordinates are $(u,v)$. I want to construct the point corresponding to $(u,-v)$ on the Edwards curve.
I can construct $(-u,-v)$ easily, that just amounts to $-nP$, flipping the $y$ coordinate. But I don't know about $(u,-v)$.
Is anything known about this construction?
See also, my related question here:
I think this amounts to a similar thing.