It turns out that the OP will be satisfied with just one form for each of these square discriminants that maps to itself. I should already say that this thing reminds me of Watson transformations. However, few_reps has shown that some of the forms of the genus interchange. It seems this mapping permutes the genus, and one might need to search for a very long time to find a case when this permutation is a derangement.

I have two examples that suggest a derangement is going to be hard to find. If prime $q \equiv 3 \pmod 4,$ make a quaternary form out of two copies of the binary $x^2 + xy + \left( \frac{q+1}{4} \right) y^2.$ This works, so half the primes are finished.

Next, if $p = 6k-1,$ take matrix $$ \left( \begin{array}{cccc} 2 & 1 & 1 & 1 \\ 1 & 2 & 0 & 1 \\ 1 & 0 & 4k & 2k \\ 1 & 1 & 2k & 4k \end{array} \right) $$ with determinant $p^2.$ The inverse times $p$ is

$$ \left( \begin{array}{rrrr} 4k & -2k & -1 & 0 \\ -2k & 4k & 1 & -1 \\ -1 & 1 & 2 & -1 \\ 0 & -1 & -1 & 2 \end{array} \right) $$ I will need to check for the explicit change of variables matrix, but it looks good.

Got it, in

$$ K = \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) $$

It turns out that the OP will be satisfied with just one form for each of these square discriminants that maps to itself. I should already say that this thing reminds me of Watson transformations. However, few_reps has shown that some of the forms of the genus interchange. It seems this mapping permutes the genus, and one might need to search for a very long time to find a case when this permutation is a derangement.

I have two (infinite sets of) examples that suggest a derangement is going to be hard to find. If prime $q \equiv 3 \pmod 4,$ make a quaternary form out of two copies of the binary $x^2 + xy + \left( \frac{q+1}{4} \right) y^2.$ This works, so half the primes are finished.

Next, if $p = 6k-1,$ take matrix $$ \left( \begin{array}{cccc} 2 & 1 & 1 & 1 \\ 1 & 2 & 0 & 1 \\ 1 & 0 & 4k & 2k \\ 1 & 1 & 2k & 4k \end{array} \right) $$ with determinant $p^2.$ The inverse times $p$ is

$$ \left( \begin{array}{rrrr} 4k & -2k & -1 & 0 \\ -2k & 4k & 1 & -1 \\ -1 & 1 & 2 & -1 \\ 0 & -1 & -1 & 2 \end{array} \right) $$ I will need to check for the explicit change of variables matrix, but it looks good.

Got it, in

$$ K = \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) $$

It turns out that the OP will be satisfied with just one form for each of these square discriminants that maps to itself. I should already say that this thing reminds me of Watson transformations. However, few_reps has shown that some of the forms of the genus interchange. It seems this mapping permutes the genus, and one might need to search for a very long time to find a case when this permutation is a derangement.

I have two examples that suggest a derangement is going to be hard to find. If prime $q \equiv 3 \pmod 4,$ make a quaternary form out of two copies of the binary $x^2 + xy + \left( \frac{q+1}{4} \right) y^2.$ This works, so half the primes are finished.

Next, if $p = 6k-1,$ take matrix $$ \left( \begin{array}{cccc} 2 & 1 & 1 & 1 \\ 1 & 2 & 0 & 1 \\ 1 & 0 & 4k & 2k \\ 1 & 1 & 2k & 4k \end{array} \right) $$ with determinant $p^2.$ The inverse times $p$ is

$$ \left( \begin{array}{cccc} 4k & -2k & -1 & 0 \\ -2k & 4k & 1 & -1 \\ -1 & 1 & 2 & -1 \\ 0 & -1 & -1 & 2 \end{array} \right) $$ I will need to check for the explicit change of variables matrix, but it looks good.

It turns out that the OP will be satisfied with just one form for each of these square discriminants that maps to itself. I should already say that this thing reminds me of Watson transformations. However, few_reps has shown that some of the forms of the genus interchange. It seems this mapping permutes the genus, and one might need to search for a very long time to find a case when this permutation is a derangement.

I have two examples that suggest a derangement is going to be hard to find. If prime $q \equiv 3 \pmod 4,$ make a quaternary form out of two copies of the binary $x^2 + xy + \left( \frac{q+1}{4} \right) y^2.$ This works, so half the primes are finished.

Next, if $p = 6k-1,$ take matrix $$ \left( \begin{array}{cccc} 2 & 1 & 1 & 1 \\ 1 & 2 & 0 & 1 \\ 1 & 0 & 4k & 2k \\ 1 & 1 & 2k & 4k \end{array} \right) $$ with determinant $p^2.$ The inverse times $p$ is

$$ \left( \begin{array}{rrrr} 4k & -2k & -1 & 0 \\ -2k & 4k & 1 & -1 \\ -1 & 1 & 2 & -1 \\ 0 & -1 & -1 & 2 \end{array} \right) $$ I will need to check for the explicit change of variables matrix, but it looks good.

Got it, in

$$ K = \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) $$