A basic application of Mackey theory This is Exercise 32.5 from the chapter on Mackey theory in the textbook Lie groups by D. Bump. Any suggestions and comments would be appreciated. Thanks!
Let $G=\left\{\begin{pmatrix}
1 & x & z \\
0 & 1 & y \\
0 & 0 & 1
\end{pmatrix}:x,y,z,\in\Bbb F_q\right\}$. Let $Z=\left\{\begin{pmatrix}
1 & 0 & z \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}:z\in \Bbb F_q\right\}$ and $A=\left\{\begin{pmatrix}
1 & 0 & z \\
0 & 1 & y \\
0 & 0 & 1
\end{pmatrix}:y,z\in \Bbb F_q\right\}$. Note that $Z$ is in the center of $G$ and $A$ is an abelian subgroup of $G$ but it contains non central elements. (So, $G\geq A\gneq Z$.) Let $\chi$ and $\psi$ be two linear characters of $A$. Assume that $\chi$ and $\psi$ have nontrivial restrictions to $Z$. Let $\chi^G$ and $\psi^G$ be the induced representations. Use Mackey theory to prove that
$$\dim\operatorname{Hom}_G(\chi^G,\psi^G)=  \begin{cases} 
   1 & \text{if $\chi,\psi$ have the same restriction to Z}  \\
   0       & \text{otherwise.}
  \end{cases}
$$

Attempt. I think we may use the following theorem.
Let $G$ be a finite group and $H_1,H_2$ be subgroups of $G$. Let $(\pi_1, V_1)$ and $(\pi_2,V_2)$ be representations of $H_1$ and $H_2$, respectively. For any $\gamma$, we define $H_\gamma=H_2\cap \gamma H_1\gamma^{-1}$ and define two representations $(\pi_1^\gamma, V_1)$ and $(\pi_2^\gamma, V_2)$ of $H_\gamma$ as follows. The representation  $\pi_2^\gamma$ is just the restriction of $\pi_2$ to $H_\gamma$. On the other hand, we define $\pi_1^\gamma(h)=\pi_1(\gamma^{-1}h\gamma)$ for $h\in H_\gamma$.
Theorem 32.2.
Let $\{\gamma_1,\gamma_2,\dots,\gamma_n\}$ be a complete set of representatives of the double cosets in $H_2\setminus G / H_1$. Then we have
$$\dim\operatorname{Hom}_G(V_1^G,V_2^G)=\sum_{i=1}^n\dim\operatorname{Hom}_{H_{\gamma_i}}(\pi_1^{\gamma_i},\pi_2^{\gamma_i}).$$
From here, if $H_1=H_2=A$ then I think we need to find a complete set of representatives of the double cosets in $H_2\backslash G / H_1.$
 A: Not only is $A$ an Abelian subgroup, but also it is a normal Abelian subgroup, since it contains $Z$, which in turn contains the commutator subgroup of $G$.  (In fact, $G$ is often presented in the exact sequence $1 \to Z \to G \to \mathbb F_q^2 \to 1$ via $\begin{pmatrix} 1 & x & z \\ & 1 & y \\ && 1 \end{pmatrix} \mapsto (x, y)$, with respect to which the commutator becomes a symplectic pairing on $\mathbb F_q^2 \times \mathbb F_q^2 \to Z \cong \mathbb F_q$.)
Thus, $A\backslash G/A$ is just $G/A$, which is $\mathbb F_q$ via $\begin{pmatrix} 1 & x & z \\ & 1 & y \\ && 1 \end{pmatrix} \mapsto x$.
It thus suffices to determine for which $x \in \mathbb F_q$ we have that, in your notation, $\chi^{\gamma_x^+}$ intertwines over $H_{\gamma_x} = A$ with $\psi^{\gamma_x^+}$—that is to say, for which $x$ they are equal—where $\gamma_x^+ = \begin{pmatrix} 1 & x \\ & 1 \\ && 1 \end{pmatrix}$.  Note that $\left[\gamma_x^+, \begin{pmatrix} 1 \\ & 1 & y \\ && 1 \end{pmatrix}\right]$ equals $\begin{pmatrix} 1 && x y \\ & 1 \\ && 1 \end{pmatrix}$ for all $x, y \in \mathbb F_q$.  If we let $x_0$ be the unique element of $\mathbb F_q$ such that $\psi\chi^{-1}$ (which is trivial on $Z$) equals
$$
\begin{pmatrix} 1 && z \\ & 1 & y \\ && 1 \end{pmatrix} \mapsto \chi\begin{pmatrix} 1 && x_0 y \\ & 1 \\ && 1 \end{pmatrix} = \psi\begin{pmatrix} 1 && x_0 y \\ & 1 \\ && 1 \end{pmatrix}
$$
(which exists because any character of $\mathbb F_q \cong A/Z$ can be obtained from any non-degenerate character of $\mathbb F_q \cong Z$ by scaling by an appropriate element of $\mathbb F_q)$, then $\chi^{\gamma_x^+}$ equals $\psi^{\gamma_x^+}$ if and only if $x$ equals $x_0$ (unless I've made a sign error, and it should be $x = -x_0$).  (The key point is that $A$ is the pullback to $G$ of a Lagrangian subspace of $\mathbb F_q^2$, i.e., a maximal subspace on which the symplectic pairing is trivial.)
Thus, in your Mackey-theoretic computation of the intertwining number, only $x = x_0$ contributes, giving $\dim \operatorname{Hom}_G(\chi^G, \psi^G) = \dim \operatorname{Hom}_A(\chi^{\gamma_{x_0}^+}, \psi^{\gamma_{x_0}^+}) = 1$.
