It is clear that a counterclockwise rotation of $90 ^\circ$ transforms $\overline {R_1Q_1}$ into $\overline{R_2Q_2}$. (In this post $\overline {R_1Q_1}$ will mean the extended line through $R_1$ and $Q_1$.) Let $\overline{R_1Q_1} \cap \overline{R_2Q_2}=B_1$, and define $A_1,C_1,D_1$ similarly. Now $\angle AB_1C = 90^\circ$ and $\overline{BB_1}$ bisects this angle. In fact a counterclockwise rotation about any point on $\overline{B_1B}$of $90 ^\circ$ transforms $\overline {R_1Q_1}$ into $\overline{R_2Q_2}$. If we can prove that $\overline{AA_1},\overline{BB_1},\overline{CC_1},\overline{DD_1}$ at some point $X$, then the counterclockwise rotation of $90 ^\circ$ about $X$ will transform $P_{1}Q_{1}R_{1}S_{1}$ into $P_{2}Q_{2}R_{2}S_{2}$. Construct $B_2$ on the opposite side of $AC$ from $B$ so that $AB_2C$ is an isosceles right triangle with vertex $B_2$. $AB_1CB_2$ is cyclic (two opposite right angles), so $\overline {B_1B_2}$ bisects $\angle AB_1C$ as well. Thus $B_1,B,B_2$ are collinear. Defining $A_2,C_2,D_2$ similarly, we have that because $\overline {AC}$ and $\overline{BD}$ are perpendicular, the squares $AB_2CD_2$ and $A_2BC_2D$ have parallel sides (and diagonals of course). Hence the lines connecting corresponding vertices are concurrent.