Let $O$ be the origin, and let $OABC$ be a rectangle, where $A$ and $C$ lie on the parabola $y = x^2.$  Then vertex $B$ must lie on a fixed parabola.  Enter the equation of the fixed parabola in the form "$y = px^2 + qx + r$".
Let $A = (a,a^2)$ and $C = (c,c^2).$  Since $\overline{OA}$ and $\overline{OC}$ are perpendicular, the product of their slopes is $-1$:
\[\frac{a^2}{a} \cdot \frac{c^2}{c} = -1.\]Thus, $ac = -1.$

[asy]
unitsize(2 cm);

real func (real x) {
  return(x^2);
}

pair A, B, C, O;

O = (0,0);
A = (0.8,func(0.8));
C = (-1/0.8,func(-1/0.8));
B = A + C - O;

draw(graph(func,-1.6,1.6));
draw(O--A--B--C--cycle);

dot("$A = (a,a^2)$", A, SE);
dot("$B$", B, N);
dot("$C = (c,c^2)$", C, SW);
dot("$O$", O, S);
[/asy]

As a rectangle, the midpoints of the diagonals coincide.  The midpoint of $\overline{AC}$ is
\[\left( \frac{a + c}{2}, \frac{a^2 + c^2}{2} \right),\]so $B = (a + c,a^2 + c^2).$

Let $x = a + c$ and $y = a^2 + c^2.$  We want a relationship between $x$ and $y$ in the form of $y = px^2 + qx + r.$  We have that
\[x^2 = (a + c)^2 = a^2 + 2ac + c^2 = a^2 + c^2 - 2 = y - 2,\]so the fixed parabola is $\boxed{y = x^2 + 2}.$