Wondering how well this would work out in $4$ by $4,$ given that David says the individual steps in the "algorithm" preserve something important:
$$ G =
\left(
\begin{array}{cccc}
2 & a & b & c \\
a & 2 & d & e \\
b & d & 2 & f \\
c & e & f& 2
\end{array}
\right)
$$
$$ \det G = a^2 f^2 + b^2 e^2 + c^2 d^2 - 2 (a b e f + a c d f + b c d e) + 4 (a b d + a c e + b c f + d e f) -4 ( a^2 + b^2 + c^2 + d^2 + e^2 + f^2 ) + 16 $$
where the degree four terms can be regarded as the indefinite ternary form $\langle 1,1,1,-2,-2,-2 \rangle$ in $x=af, y=be,z=cd.$
$$ S =
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & f \\
0 & 0 & 0& -1
\end{array}
\right)
$$
$$ S^t G S =
\left(
\begin{array}{cccc}
2 & a & b & fb-c \\
a & 2 & d & fd-e \\
b & d & 2 & f \\
fb-c & fd-e & f& 2
\end{array}
\right)
$$
So, this basic operation (an involution) flips two positions at once..
$$ T =
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & f& 1
\end{array}
\right)
$$
$$ T^t G T =
\left(
\begin{array}{cccc}
2 & a & fc-b & c \\
a & 2 & fe-d & e \\
fc-b & fe-d & 2 & f \\
c & e & f& 2
\end{array}
\right)
$$
hmmmm.......
Just checking,
$$ N_1 =
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0& -1
\end{array}
\right)
$$
$$ N_1 G N_1 =
\left(
\begin{array}{cccc}
2 & a & b & -c \\
a & 2 & d & -e \\
b & d & 2 & -f \\
-c & -e & -f& 2
\end{array}
\right)
$$
$$ N_2 =
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0& -1
\end{array}
\right)
$$
$$ N_2 G N_2 =
\left(
\begin{array}{cccc}
2 & a & -b & -c \\
a & 2 & -d & -e \\
-b & -d & 2 & f \\
-c & -e & f& 2
\end{array}
\right)
$$
Right, that is enough to know on these because $-I$ changes nothing.
Really ought to see what a permutation, a single transposition, does
$$ T_2 =
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{array}
\right)
$$
$$ T_2 G T_2 =
\left(
\begin{array}{cccc}
2 & a & c & b \\
a & 2 & e & d \\
c & e & 2 & f \\
b & d & f& 2
\end{array}
\right)
$$
Swaps two pairs