= Solution
Write a homogeneous <quadratic polynomial> as $p(x,y)=ax^2+bxy+cy^2$. Invariance under $B:(x,y)\mapsto(x,-y)$ forces $b=0$. Invariance under $A:(x,y)\mapsto(y,x)$ then forces $a=c$. Since $A,B$ generate the <group>, this proves that all homogeneous quadratic <polynomial invariants> are
$$
\boxed{p(x,y)=a(x^2+y^2),\qquad a\in\mathbb C.}
$$
This agrees with the single <trivial representation> in the <symmetric square> in <1/d/solution>; more precisely, coordinate <polynomials> lie in the <symmetric square> of the <dual representation>, which here is equivalent to $\rho$. If “quadratic” permits terms of smaller degree, arbitrary constants can be added. There are no invariant linear terms because $-I\in G$ negates every linear <polynomial>.
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