Proving Diamond & Shurman Exercise 3.7.1, about conjugacy class of $\Gamma_0^{\pm}(N)$. I am reading chapter 3 of A First Course in Modular Forms but have troubles in Exercise 3.7.1 (c) and (d).

(c) Show that the $\Gamma_0^{\pm}(N)$-conjugacy class of $\gamma \in \Gamma_0(N)$ is the union of the $\Gamma_0(N)$ conjugacy classes of $\gamma$ and $\begin{bmatrix}  1 & 0 \\ 0 & -1 \end{bmatrix} \gamma \begin{bmatrix}  1 & 0 \\ 0 & -1 \end{bmatrix}$. Show that if $\gamma$ has order $4$ or $6$ then this union is disjoint.


(d) Let $\gamma = \begin{bmatrix}  1 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix}  0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix}  1 & 1 \\ 1 & 2 \end{bmatrix}^{-1} = \begin{bmatrix}  3 & -2 \\ 5 & -3 \end{bmatrix}$, an order-$4$ element of $\Gamma_0(5)$. Show that $\gamma$ is not conjugate to its inverse in $\Gamma_0(5)$.

Here let $\mathrm{GL}_2(\mathbb{Z})$ be the group of invertible $2 \times 2$ matrices with integer entries, $\mathrm{SL}_2(\mathbb{Z})$ the group of $2 \times 2$ matrices with integer entries and determinant $1$, and
\begin{align*}
\Gamma_0^{\pm}(N) &= \left\{ \begin{bmatrix}  a & b \\ c & d \end{bmatrix} \in \mathrm{GL}_2(\mathbb{Z}) : \begin{bmatrix}  a & b \\ c & d \end{bmatrix} \equiv \begin{bmatrix}  * & * \\ 0 & * \end{bmatrix} \pmod{N} \right\} \\
\Gamma_0(N) &= \left\{ \begin{bmatrix}  a & b \\ c & d \end{bmatrix} \in \mathrm{SL}_2(\mathbb{Z}) : \begin{bmatrix}  a & b \\ c & d \end{bmatrix} \equiv \begin{bmatrix}  * & * \\ 0 & * \end{bmatrix} \pmod{N} \right\}.
\end{align*}
I have proved the first part of (c): the map
\begin{align*}
\left(\Gamma_0^{\pm}(N) \setminus \Gamma_0(N)\right) &\longrightarrow \Gamma_0(N) \\
\alpha &\longmapsto \alpha \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}
\end{align*}
is a bijection, so
\begin{align*}
\left\{ \beta \gamma \beta^{-1}: \beta \in \Gamma_0^{\pm}(N) \setminus \Gamma_0(N) \right\} 
= \left\{ \alpha \begin{bmatrix}  1 & 0 \\ 0 & -1 \end{bmatrix} \gamma \begin{bmatrix}  1 & 0 \\ 0 & -1 \end{bmatrix} \alpha^{-1}: \alpha \in \Gamma_0(N) \right\}.
\end{align*}
How to show the remaining parts? For part (d), I think one cannot apply (c) directly because (d) serves as an example of the following statement in P.93

The extended conjugacy class of $\gamma$ under $\Gamma_0^{\pm}(N)$ is not in general the union of the conjugacy class of $\gamma$ and $\gamma^{-1}$ under $\Gamma_0(N)$.

 A: Let
$$\Gamma_0^-(N)=\left\{ \begin{bmatrix}
 a & b\\
 c & d
\end{bmatrix}\in\Gamma_0^\pm(N):ad-bc=-1 \right\},$$
so the $\Gamma_0^\pm(N)$-conjugacy classes of $\gamma$ is the union of $\Gamma_0(N)$-conjugacy classes of $\gamma$ and $\Gamma_0^-(N)$-conjugacy classes of $\gamma$. Let $\alpha=\begin{bmatrix}
 a & b\\
 c & d
\end{bmatrix}\in\Gamma_0^-(N)$, then
$$\alpha\gamma\alpha^{-1}=\begin{bmatrix}
 a & -b\\
 c & -d
\end{bmatrix}\begin{bmatrix}
 1 & 0\\
 0 & -1
\end{bmatrix}\gamma\begin{bmatrix}
 1 & 0\\
 0 & -1
\end{bmatrix}\begin{bmatrix}
 -d & b\\
 -c & a
\end{bmatrix},$$
where $\begin{bmatrix}
 a & -b\\
 c & -d
\end{bmatrix}\in\Gamma_0(N)$ and $\begin{bmatrix}
 -d & b\\
 -c & a
\end{bmatrix}=\begin{bmatrix}
 a & -b\\
 c & -d
\end{bmatrix}^{-1}$. Let $\alpha=\begin{bmatrix}
 a & b\\
 c & d
\end{bmatrix}\in\Gamma_0(N)$ this time, then
$$\alpha\begin{bmatrix}
 1 & 0\\
 0 & -1
\end{bmatrix}\gamma\begin{bmatrix}
 1 & 0\\
 0 & -1
\end{bmatrix}\alpha^{-1}=\begin{bmatrix}
 a & -b\\
 c & -d
\end{bmatrix}\gamma\begin{bmatrix}
 d & -b\\
 c & -a
\end{bmatrix},$$
where $\begin{bmatrix}
 a & -b\\
 c & -d
\end{bmatrix}\in\Gamma_0^-(N)$ and $\begin{bmatrix}
 d & -b\\
 c & -a
\end{bmatrix}=\begin{bmatrix}
 a & -b\\
 c & -d
\end{bmatrix}^{-1}$.
Thus the $\Gamma_0^-(N)$-conjugacy classes of $\gamma$ is equal to the $\Gamma_0(N)$-conjugacy classes of $\begin{bmatrix}
 1 & 0\\
 0 & -1
\end{bmatrix}\gamma\begin{bmatrix}
 1 & 0\\
 0 & -1
\end{bmatrix}$. The first statement of (c) is true.
By Proposition 2.3.3, if $\gamma$ has order $4$ then $\gamma$ is conjugate to $\begin{bmatrix}
 0 & -1\\
 1 & 0
\end{bmatrix}^{\pm1}=\pm\begin{bmatrix}
 0 & -1\\
 1 & 0
\end{bmatrix}$ in $\mathrm{SL}_2(\mathbb Z)$. If this two conjugacy classes intersects, i.e. $\gamma$ is conjugate to $\begin{bmatrix}
 1 & 0\\
 0 & -1
\end{bmatrix}\gamma\begin{bmatrix}
 1 & 0\\
 0 & -1
\end{bmatrix}$ in $\Gamma_0(N)$, then $\gamma$ is conjugate to $\gamma$ in $\Gamma_0^-(N)$, then $\pm\begin{bmatrix}
 0 & -1\\
 1 & 0
\end{bmatrix}$ is conjugate to $\pm\begin{bmatrix}
 0 & -1\\
 1 & 0
\end{bmatrix}$ in $\mathrm{GL}_2^-(\mathbb Z)$, where
$$\mathrm{GL}_2^-(\mathbb Z)=\left\{ \begin{bmatrix}
 a & b\\
 c & d
\end{bmatrix}\in\mathrm{GL}_2(\mathbb Z):ad-bc=-1 \right\}.$$
Assuming $\pm\begin{bmatrix}
 0 & -1\\
 1 & 0
\end{bmatrix}=\pm\alpha\begin{bmatrix}
 0 & -1\\
 1 & 0
\end{bmatrix}\alpha^{-1}$ and $\alpha=\begin{bmatrix}
 a & b\\
 c & d
\end{bmatrix}\in\mathrm{GL}_2^-(\mathbb Z)$, then
$$\pm\begin{bmatrix}
 0 & -1\\
 1 & 0
\end{bmatrix}=\pm\begin{bmatrix}
 -(ac+bd) & a^2+b^2\\
 -(c^2+d^2) & ac+bd
\end{bmatrix}.$$
Thus $a^2+b^2=-1$, a contradiction.
If $\gamma$ has order $6$, we deduce that $\begin{bmatrix}
 0 & -1\\
 1 & 1
\end{bmatrix}$ and $\begin{bmatrix}
 1 & 1\\
 -1 & 0
\end{bmatrix}$ are conjugate to themselves in $\mathrm{GL}_2^-(\mathbb Z)$ by similar way, and which yields
$$\begin{bmatrix}
 0 & -1\\
 1 & 1
\end{bmatrix}=\begin{bmatrix}
 bc-ac-bd & a^2+b^2-ab\\
 cd-c^2-d^2 & ac+bd-ad
\end{bmatrix},$$
$$\begin{bmatrix}
 1 & 1\\
 -1 & 0
\end{bmatrix}=\begin{bmatrix}
 ac+bd-ad & ab-a^2-b^2\\
 c^2+d^2-cd & bc-ac-bd
\end{bmatrix}.$$
Thus we have $\left( a-\frac12b \right)^2+\frac34b^2=-1$ and $\left( c-\frac12d \right)^2+\frac34d^2=-1$, contradictions.
