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 

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