[This is more of a comment to Franz Lemmermeyer's answer.]

Let $F(x,y)=ax^2+bxy+cy^2=[a,b,c]$ be an integral ($a,b,c\in\mathbb{Z}$) binary quadratic form of discriminant $\Delta=b^2-4ac$, and let $\mathcal{Q}=\mathcal{Q}(\Delta)$ be the collection of all such forms. $GL_2(\mathbb{Z})\times\langle i\rangle$ acts on $\mathcal{Q}$ by change of variable, preserving the discriminant
$$
g=\left(\begin{array}{cc}\alpha&\beta\\\gamma&\delta\end{array}\right), \ F^g=F(\alpha x+\beta y,\gamma x+\delta y).
$$
where
$$
i=\left(\begin{array}{cc}\sqrt{-1}&0\\0&\sqrt{-1}\\\end{array}\right)
$$
(which has determinant $-1$ and preserves integrality $[a,b,c]^i=[-a,-b,-c]$).

Consider the actions of
$$
s=\left(\begin{array}{cc}0&-\sqrt{-1}\\\sqrt{-1}&0\\\end{array}\right), \ [a,b,c]^s=[-c,b,-a], \ s^2=1
$$
and
$$
f=\left(\begin{array}{cc}1&1\\0&-1\\\end{array}\right), \ [a,b,c]^{f}=[a,2a-b,a-b+c], \ f^2=1.
$$
If $\mathcal{Q}$ contains a fixed point for the action of $s$, (i.e. $[a,b,-a]$), then $\Delta=b^2+(2a)^2$ is a sum of two squares, and if $\mathcal{Q}$ contains a fixed point for the action of $f$, (i.e. $[a,a,c]$), then $\Delta=a(a-4c)$ factors.

One interesting aspect of these involutions is that they go outside $SL_2(\mathbb{Z})$ to the broader orthogonal group $O(\Delta;\mathbb{Z})$ preserving the discriminant,
$$
\Delta(a,b,c)=\left(a \ b \ c\right)
\left(
\begin{array}{ccc}
0&0&-2\\
0&1&0\\
-2&0&0\\
\end{array}
\right)
\left(
\begin{array}{c}
a\\
b\\
c\\
\end{array}
\right).
$$

Other than that, they are among the ``simplest'' involutions available.

Now, if $\Delta=p>0$ is prime, then $f$ has fixed points only when $p\equiv 1\bmod 4$, namely
$$
\pm\left[1,1,\frac{1-p}{4}\right], \ \pm\left[p,p,\frac{p-1}{4}\right].
$$
One might hope to finagle a finite set of ``reduced'' forms that contains only one of these fixed points and is closed under both involutions, but this seems somewhat complicated (or impossible?). [For instance, looking at $s$ with uniqueness in mind, one might consider those forms with $a>0>c$. Considering what happens with these under $f$, we get the condition $a+c<b$. Looking at this condition under $s$, we get $|a+c|<b$ (so $b>0$) and $b>0$ implies $b<2a$ for stability under $f$, etc.]

The other two elements from $O(\Delta)$ (acting from the left on the column $(a,b,c)$ from my own predjudices) in the ``one-sentence'' proof aren't involutions; they're of order 6 and inverse to one another:
$$
\left(
\begin{array}{ccc}
0&0&-1\\
0&1&-2\\
-1&1&-1\\
\end{array}
\right)\leftrightarrow
\left(\begin{array}{cc}0&\sqrt{-1}\\-\sqrt{-1}&-\sqrt{-1}\\\end{array}\right)=t,
$$
$$
\left(
\begin{array}{ccc}
-1&1&-1\\
-2&1&0\\
-1&0&0\\
\end{array}
\right)\leftrightarrow
\left(\begin{array}{cc}\sqrt{-1}&-\sqrt{-1}\\-\sqrt{-1}&0\\\end{array}\right)=t^{-1}
$$
i.e.
\begin{align*}
[a,b,c]^t&=[-c,b-2c,-a+b-c],\\
[a,b,c]^{t^{-1}}&=[-a+b-c,-2a+b,-a],
\end{align*}
under the association
$$
F^g\longleftrightarrow
\left(
\begin{array}{ccc}
\alpha^2&\alpha\gamma&\gamma^2\\
2\alpha\beta&\alpha\delta+\beta\gamma&2\gamma\delta\\
\beta^2&\beta\delta&\delta^2\\
\end{array}
\right)
\left(
\begin{array}{c}
a\\
b\\
c\\
\end{array}
\right).
$$
Again, these are some of the simpler finite order $O(\Delta)$ elements (from the $2\times2$ matrix standpoint).

Now, what happens to the finite set of ``reduced'' forms
$$
\mathcal{Q}_0=\{[a,b,c]\in\mathcal{Q} : a>0>c, b>0\}
$$
under $t$ and $t^{-1}$? We have
\begin{align*}
[a,b,c]^t&\in\mathcal{Q} \Longleftrightarrow b<a+c,\\
[a,b,c]^{t^{-1}}&\in\mathcal{Q} \Longleftrightarrow b>2a \text{ and } b>a+c.
\end{align*}
These conditions seem to start matching up with the the earlier attempt to find reduced forms stabalized by both $s$ and $f$. So much so that the map
$$
[a,b,c]\mapsto\left\{
\begin{array}{cc}
[a,b,c]^t & b<a+c\\
[a,b,c]^f& a+c<b<2a\\
[a,b,c]^{t^{-1}} & b>2a\\
\end{array}
\right.
$$
is an involution on $\mathcal{Q}_0$ giving us what we want!

2more comments