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