Take the simple system of parametric equations, $$x=\sin(at)$$ $$y=\sin(bt)$$ where $a,b \in \Bbb{N}$.
When is this curve symmetric with respect to both the $x$ and $y$ axes? In other words, what values of $a$ and $b$ satisfy the condition that for every point on the curve $P_1:(x, y)$, there also exist $P_2:(x, -y), P_3:(-x, y), P_4:(-x, -y)$?
Assume $a,b >0$ and that $a/b$ is rational. Then by rescaling one may assume $a,b$ are coprime positive integers. We claim the symmetry occurs provided $a,b$ have opposite parity. Suppose e.g. $a=2s,b=2r-1,$ and consider comparing $(x(t),y(t))$ with $x(t+\pi),y(t+\pi).$ Note that $2s(t+\pi - t)=2s\pi,$ so that $x(t)=x(t+\pi).$ On the other hand, on comparing $y(t)$ with $y(t+\pi)$ we have $(2r-1)(t+\pi - t)=(2r-1)\pi,$ an odd multiple of $\pi,$ so that $y(t+\pi)=-y(t).$
This shows each point $(x,y)$ on the graph has also the point $(x,-y)$ on the graph, and then the other symmetries follow (as noted by copper.hat's comment) by $\sin$ being an odd function.
Note this is only a partial answer in that I didn't prove the symmetries do not hold when $a,b$ have the same parity and $\gcd$ 1, i.e. when $a,b$ are odd.
