The Brahmagupta-Fibonacci identity states that the set of the sum of two squares is closed under multiplication. $$(u^2 + v^2)(A^2 + B^2) = (uA \pm vb)^2 + (vA \mp uB)^2$$ This is easy to verify as \begin{equation} \begin{split} (uA + vb)^2 + (vA - uB)^2\\ & = (u^2A^2 + 2uAvB + v^2B^2) + (v^2A^2 -2vAuB + u^2b^2)\\ & = (u^2A^2 + v^2B^2 + v^2A^2 + u^2B^2) \\ & = (u^2 + v^2)(A^2 + B^2) \end{split} \end{equation} The plus and minus can be flipped in the result.
However, suppose the case where $u \bot v \land A \bot B$ (u and v are relatively prime and A and B are relatively prime). I am pretty sure this should imply that $(uA + vB) \bot (vA - uB)$. Provided that $(u^2 + v^2) \land (A^2 + B^2)$ are odd, so that the product must be odd. however I'm not so sure how to go about it.
I know that according to the extended Euclidean algorithm,
$$ ux_1 + vx_2 = 1$$ and $$ Ax_3 + Bx_4 = 1$$
But I cant seem to get that to imply anything about the two quantities of $(uA + vB)$ and $(vA - uB)$. Are there some other tricks of the gcd to make this work?
EDIT:
This conjecture needs to narrow down its preconditions. The product $(u^2 + v^2)(A^2 + B^2) = (uA \pm vb)^2 + (vA \mp uB)^2$ must be odd. Also, because there are two ways this new number can be written as the sum of two squares, I still claim that at least one of these pairs will be squares of relatively prime numbers.
It does not. As a counterexample, consider $$(u,v) = (51,73), \quad (A,B) = (37,41),$$ both pairs of which are relatively prime. Their sums of squares is $$u^2 + v^2 = 7930, \quad A^2 + B^2 = 3050,$$ and the product is $24186500$, which also happens to be $(uA + vB)^2 + (vA - uB)^2 = 4880^2 + 610^2$. But $\gcd(4880, 610) = 610$. There are smaller counterexamples but this one seemed interesting since in fact $4880 = 8 \cdot 610$.
I found another counterexample to satisfy all your additional criteria. You require that the product $(u^2+v^2)(A^2+B^2)$ is odd, and at least one of the two ways this product can be written as the sum of two squares according to the formula will consist of a pair of relatively prime integers; however, $$(u,v) = (1,8), \quad (A,B) = (2, 29)$$ gives the product $$(u^2+v^2)(A^2+B^2) = 54925 = 234^2 + 13^2 = 230^2 + 45^2,$$ yet $$\gcd(234,13) = 13, \quad \gcd(230, 45) = 5.$$