I know this not, everybody else seems to. This is from page 215 of Cassels, Rational Quadratic Forms, formula 4.1, or SPLAG, page 389 formula (36). quote:

If $f$ and $g$ are forms of determinant $d$ in the same genus, then they are rationally equivalent by some transformation whose denominator is prime to $2d.$ Hence we can find corresponding lattices $L,M$ for which $$ [ L : L \cap M ] = [ M : L \cap M] > = r, $$ say, for some number $r$ which is prime to $2d.$

We are talking about Siegel's definition of forms being in the same genus if they are **rationally equivalent without essential denominator.**

So, here is an example in matrix slang. Given quadratic forms with symmetric matrices $$ F \; = \; \left( \begin{array}{cccc} 2 & 1 & 0 & 1 \\\ 1 & 2 & 0 & 0 \\\ 0 & 0 & 2 & 1 \\\ 1 & 0 & 1 & 6 \end{array} \right) $$ and $$ G \; = \; \left( \begin{array}{cccc} 2 & 1 & 1 & 0 \\\ 1 & 2 & 0 & 1 \\\ 1 & 0 & 2 & 0 \\\ 0 & 1 & 0 & 8 \end{array} \right) $$ Next, I take $r=3,$ because the discriminant is $29,$ and find $$ P \; = \; \left( \begin{array}{cccc} 1 & 1 & 1 & 7 \\\ 1 & -2 & 1 & -5 \\\ 1 & 1 & -2 & 1 \\\ 1 & 1 & 1 & -2 \end{array} \right) $$ that satisfies $$ P^T \; F \; P \; = \; 9 \; G = r^2 \; G. $$

Now, here is the trick. To get back from $G$ to $F,$ it appears that one needs to use $Q = \; \mbox{adj} \; P$ which has a much larger determinant, so things look asymmetric. Indeed, $$ Q \; = \; \; \mbox{adj} \; P \; = \; \left( \begin{array}{cccc} -18 & -27 & -27 & -9 \\\ 9 & 27 & 0 & -36 \\\ -9 & 0 & 27 & -18 \\\ -9 & 0 & 0 & 9 \end{array} \right) $$ However, a miracle! The GCD of these entries is 9, and we get the improved $$ Q_1 \; = \; \left( \begin{array}{cccc} -2 & -3 & -3 & -1 \\\ 1 & 3 & 0 & -4 \\\ -1 & 0 & 3 & -2 \\\ -1 & 0 & 0 & 1 \end{array} \right). $$ The we really do get what we wanted, $$ P Q_1 = -9 I = \pm r^2 I $$ and $$ Q_1^T \; G \; Q_1 \; = \; 9 \; F = \; r^2 F. $$

Alright, so here is the **question,** with the dimension $n$ thrown in: Suppose $F,G$ are symmetric positive definite matrices of integers with the same determinant $d.$ Suppose we have an integer $r$ with $\gcd (r, 2 d) = 1.$ Suppose that we have a matrix $P$ of integers, with $\det P = \pm r^n,$ such that $ P^T \; F \; P \; = \; r^2 \; G. $ Take $Q = \; \mbox{adj} \; P,$ so that $\det Q = \pm r^{n^2 - n}$ and $PQ = QP = (\det P) I = \pm r^n I.$ Is it always the case that
$$ \gcd Q = r^{n-2} ?$$

I think this is progress. Since 1994, it is only 18 years that I have been completely confused on this point and unaware that I was confused.

EDIT: The matrix $P$ is not necessarily rank $1 \pmod r.$ Here are the two forms by Schiemann of discriminant 1729, not equivalent but in the same genus and, wait for it, the same theta series.

$$ F \; = \; \left( \begin{array}{cccc} 4 & 1 & 0 & 1 \\\ 1 & 8 & 1 & 3 \\\ 0 & 1 & 8 & 4 \\\ 1 & 3 & 4 & 10 \end{array} \right) $$ and $$ G \; = \; \left( \begin{array}{cccc} 4 & 2 & 1 & 1 \\\ 2 & 8 & -2 & 1 \\\ 1 & -2 & 10 & 5 \\\ 1 & 1 & 5 & 10 \end{array} \right) $$ Next, I take $r=5,$ and find $$ P \; = \; \left( \begin{array}{cccc} -1 & 6 & -4 & 0 \\\ -2 & -3 & -3 & 0 \\\ 3 & 2 & -3 & 0 \\\ -1 & -1 & 0 & -5 \end{array} \right) $$ that satisfies $$ P^T \; F \; P \; = \; 25 \; G = r^2 \; G. $$ We still wind up, in the same manner, with a very pleasant $$ Q_1 \; = \; \left( \begin{array}{cccc} -3 & -2 & 6 & 0 \\\ 3 & -3 & -1 & 0 \\\ -1 & -4 & -3 & 0 \\\ 0 & 1 & -1 & -5 \end{array} \right). $$