jmc

Since $y=x$ is an axis of symmetry, if point $(a,b)$ lies on the graph, then so does $(b,a).$ Thus, the equation of the graph can also be written as $$x=\frac{py+q}{ry+s}.$$Substituting $y=\frac{px+q}{rx+s},$ we get $$x=\frac{p\cdot \frac{px+q}{rx+s}+q}{r\cdot \frac{px+q}{rx+s}+s}=\frac{p(px+q)+q(rx+s)}{r(px+q)+s(rx+s)}.$$Cross-multiplying, we get $$x[r(px+q)+s(rx+s)]=p(px+q)+q(rx+s).$$Expanding, we get $$(pr+rs)x^2+(s^2-p^2)x-(pq+qs)=0.$$We can take out a factor of $p+s$: $$(p+s)(r{x}^2+(s-p)x-q)=0.$$This equation must holds for all $x.$ Since $r\ne 0,$ the quadratic $r{x}^2+(s-p)x-q$ cannot be 0 for all $x,$ so we must have $p+s=0.$ The correct statement is $\boxed{\text{(C)}}.$