About integral binary quadratic forms fixed by $\operatorname{GL_2(\mathbb Z)}$ matrices of order $3$ I am reading this paper of Manjul Bhargava and Ariel Shnidman, and I want to prove this claim, which appear at the first paragraph of Theorem $14$:

Up to $\operatorname{SL_2}(\mathbb Z)$ equivalence and scaling, there is only one integral binary quadratic form having an $\operatorname{SL_2}(\mathbb Z)$-automorphism of order three, namely $Q(x,y)=x^2+xy+y^2$.

Now the pertinent definitions in order to understand the problem. 
Consider the following action of $\operatorname{GL_2}(\mathbb Z)$ on the set of integral binary quadratic forms: for $\gamma\in$$\operatorname{GL_2}(\mathbb Z)$ and $f(x,y)=Px^2+Qxy+Ry^2$, with $P,Q,R\in\mathbb Z$, we define $(\gamma f)(x,y)=f\bigl((x,y)\gamma\bigr)$. Identifying each such $f$ with the triplet $(P,Q,R)^t$, we have that the triplet associated to $\gamma f$ is given by the matricial product
$$M_{\gamma}\,\cdot\begin{bmatrix} P\\ Q\\ R\end{bmatrix},\ \text{where}\ \
\gamma=\begin{bmatrix}
A & B\\
C & D
\end{bmatrix}\,\ \text{and}\ M_{\gamma}=\begin{bmatrix}
A^2 & AB    & B^2\\
2AC & AD+BC & 2BD\\
C^2 & CD    & D^2
\end{bmatrix}.$$
With these notations the authors' claim can be restated as follows:

Suppose that $f=(P,Q,R)^t$ is fixed by $\gamma\in\operatorname{GL_2}(\mathbb Z)$ of order $3$ and $Q\ne0$. Show that there exist $\theta\in\operatorname{GL_2}(\mathbb Z)$ and $n\in\mathbb Z$ such that $\theta f=(n,n,n)^t.$

I have had a very bad time trying to prove this fact. Perhaps my understanding of the problem is wrong, but I honestly don't think so (I successfully completed the previous details of the paper).
My work so far: Let $a=A+D$ be the trace of $\gamma$. Since $\gamma^3=I$ then $\det\gamma=1$, so the characteristic polynomial of $\gamma$ is $X^2-aX+1$. Now
$$X^3-1=(X+a)(X^2-aX+1)+(a+1)\bigl[(a-1)X-1\bigr]\,,$$
and since $\gamma$ also satisfies $\gamma^3-I=0$, then after evaluating the equality above at $\gamma$ we obtain that either $a=-1$ or $(a-1)\gamma=I$; but $(a-1)\gamma=I$ implies $(a-1)^3=1$, that is $a=2$, and so $I=(a-1)\gamma=\gamma$, contradicting the fact that $\gamma$ has order $3$ in $\operatorname{GL_2}(\mathbb Z)$.
On the other hand $\gamma^{-1}$ also fixes $f$, so denoting $(P,Q,R)^t$ by $v$ we obtain $(M_{\gamma^{-1}}-M_\gamma)v=0$. Using the formula $\gamma^{-1}=\binom{\ \ D\ \ -B}{\!\!\!-C\ \ \ \ A}$ together with the equalities $AD-BC=1$ and $A+D=-1$ we get
$$\begin{align*}
M_{\gamma^{-1}}-M_{\gamma}=&\,\begin{bmatrix}
D^2  & -BD   &  B^2\\
-2CD & AD+BC & -2AB\\
C^2  & -AC   &  A^2
\end{bmatrix}-\begin{bmatrix}
A^2 & AB    & B^2\\
2AC & AD+BC & 2BD\\
C^2 & CD    & D^2
\end{bmatrix}
\\[5mm]
=&\,\begin{bmatrix}
A-D & B & 0 \\
2C  & 0 & 2B\\
 0  & C & D-A
\end{bmatrix}\,,
\end{align*}$$
and so $v$ has the specific form $v=\frac Q{A-D}(-B,A-D,C)^t$.
Now I am stuck at this point: If $\theta=\binom{B\ \ \ 0}{D\ \ \ 1}$ then $M_\theta\cdot(n,n,n)^t=v$, where $n=\frac{-Q}{B(A-D)}$ (it is easy to see from the equalities $AD-BC=1$ and $A+D=-1$ that $B(A-D)\ne0$). The problem is, of course, that $n$ is not necessarily an integer and that $\theta$ not necessarily belong to $\operatorname{GL_2}(\mathbb Z)$. Changing $B$ by $1$ in $\theta$ does not work, because scaling the first row of $\theta$ amounts to a corresponding quadratic scaling at the first row of $M_\theta$, but keeping the same scaling at the second row of $M_\theta$, so the desired equality is not preserved.
I tried every possible matrix obtained from $\gamma$ with no success, so any help is welcomed!!!
 A: The modular group $SL_2 \mathbb Z$ is presented by
$$  \langle T,P | T^4 = P^3 = 1 \rangle,  $$
where we mean specific matrices
$$ T =
\left(
\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}
\right)
  $$
and
$$ P =
\left(
\begin{array}{rr}
0 & 1 \\
-1 & -1
\end{array}
\right)
  $$
This is from page 9 in Binary Quadratic Forms by Duncan A. Buell. Proof takes a few pages.
Now, $P$ really is in the automorphism group for $x^2 + xy + y^2,$ since
$$
\left(
\begin{array}{rr}
0 & -1 \\
1 & -1
\end{array}
\right)
\left(
\begin{array}{rr}
2 & 1 \\
1 & 2
\end{array}
\right)
\left(
\begin{array}{rr}
0 & 1 \\
-1 & -1
\end{array}
\right)=
\left(
\begin{array}{rr}
2 & 1 \\
1 & 2
\end{array}
\right)
$$
Next, I believe already (confirmed by studiosus, comment after answer as well as a separate answer) that any group element of order 3 is conjugate to $P$ or $P^2,$ of the form $gP g^{-1}$ for example. But then we simply absorb $g$ into the Gram/Hessian matrix for the form; that is, if
$$ (g^{-1})^TP^T g^T H gP g^{-1} = H,  $$ then
$$  P^T g^T H g P =  g^T H g,   $$ we rename
$$ G =  g^T H g  $$ to arrive at
$$   P^T G P =  G.  $$
Note that $g^T$ and $g^{-1}$ are not quite the same, and the expression $g^T H g$ is the correct expression for the Gram matrix of a new binary quadratic form; this new form is actually called "equivalent" to the original.
Finally, what things have $P$ in the automorphism group?
$$
\left(
\begin{array}{rr}
0 & -1 \\
1 & -1
\end{array}
\right)
\left(
\begin{array}{rr}
2A & B \\
B & 2C
\end{array}
\right)
\left(
\begin{array}{rr}
0 & 1 \\
-1 & -1
\end{array}
\right)=
\left(
\begin{array}{cc}
2C & 2C-B \\
2C-B & 2A-2B+2C
\end{array}
\right)
$$
which give the equations
$$   2A = 2C, \; \; 2C-B = B, \; \; 2A-2B+2C = 2C,   $$
so actually $A=B=C.$
As far as I can tell that is it. It would be nice if someone with a more sure hand with generators and relations confirmed, or corrected, the part about elements of order 3 in the group.
A: It is a bit too long for a comment, it provides details for Will's answer: Consider a group $G$ which splits as a free product $G_1 * G_2$. 
Theorem. Every finite order element is conjugate either into $G_1$ or into $G_2$. 
Proof. The trick is to think of this not as an algebraic but a geometric problem. The free product decomposition of $G$ corresponds to a simplicial action of $G$ on a Bass-Serre tree $T$, which is a certain simplicial tree such that the vertex stabilizers are conjugates of $G_1$ and $G_2$ (there are exactly two $G$-orbits of vertices), and the edge-stabilizers are trivial. 
The construction of this tree is not hard at all, especially if you are a topologist familiar with van Kampen theorem (and Serre does not assume any knowledge of topology and proves everything from scratch by imitating Tits' BN pairs theory). As a topologist, I would consider a wedge $K$ of $K_1=K(G_1,1)$ and $K_2=K(G_2,1)$ (call $e$ the point where they are joined together) and look at its universal cover. Lifts of $K_1$ and $K_2$ are the vertices of the tree and lifts of $e$ are the edges. By van Kampen's theorem, $\pi_1(K)\cong G_1* G_2$. Everything else is elementary. 
Now, if you have any finite-order element $g\in G$, its orbit in the vertex set of $T$ is finite and, hence, spans a finite subtree. It is an elementary argument nicely explained by Serre in his book that each group acting on a finite tree $S$ has a fixed point.  (Just keep removing all leaves of the  tree $S$ inductively until you have a fixed vertex left.) Therefore, such $g$ is conjugate into one of the subgroups $G_1$ or $G_2$. 
