A basic application of Mackey's theorem Let $G=GL(2,\mathbb {F}_q)$ and $B=\left\{\begin{pmatrix}
* & *  \\
0 & * 
\end{pmatrix}\right\}$ the
Borel subgroup of upper triangular matrices. Let $\chi_1$ and $\chi_2$ be the characters of $\mathbb{F}_q^\times$. Define the character $\chi:B\to\mathbb{C}^\times$ by $$\chi\left(\begin{pmatrix}
y_1 & *  \\
0 & y_2 
\end{pmatrix}\right)=\chi_1(y_1)\chi_2(y_2)\qquad \text{for all $y_1,y_2\in\mathbb{F}_q$ with $y_1y_2\neq 0$}.$$
Similarly, let $\mu_1,\mu_2$ be two other characters of $\mathbb{F}_q^\times$ and let $\mu:B\to\mathbb{C}^\times$ be the corresponding character of $B$ defined by $$\mu\left(\begin{pmatrix}
y_1 & *  \\
0 & y_2 
\end{pmatrix}\right)=\mu_1(y_1)\mu(y_2)\qquad \text{for all $y_1,y_2\in\mathbb{F}_q$ with $y_1y_2\neq 0$}.$$
Question. We want to prove that $Ind_B^G(\chi)\simeq Ind_B^G(\mu)$ if and only if either $\chi_1=\mu_1$ and $\chi_2=\mu_2$ or $\chi_1=\mu_2$ and $\chi_2=\mu_1$ by using Theorem 32.1 (Mackey's theorem, geometric version) from the textbook Lie groups by D.Bump.
Attempt.
The theorem implies that $Hom_G(Ind_B^G\chi,Ind_B^G\mu)$ is isomorphic as a vector space to the space, $V$, of all functions $f:G\to\mathbb {C}$ such that $f(b_2gb_1)=\mu(b_2)f(g)\chi(b_1)$ for all $b_1,b_2\in B,g\in G.$
Observation 1. Let $f\in V.$ Then, for any $b\in B$, we have \begin{align*}
 f(b)&= f(b.1)= \mu(b)f(1)    \\
 &= f(1.b)=f(1)\chi(b) \implies  0=f(1)\big(\mu(b)-\chi(b)\big). 
 \end{align*}So, if there exists a function $f\in V$ with $f(1)\neq 0$ then $\mu=\chi.$ Conversely, If $\mu=\chi$ then there exists a function $f\in V$ with $f(1)\neq 0$.
Observation 2. Note that $G=B\sqcup B\omega B$ where
$\omega = \begin{pmatrix}
0 & 1  \\
1 & 0 
\end{pmatrix}$. Let $f\in V$. For any $t\in T$, say $t=\begin{pmatrix}
y_1 & 0  \\
0 & y_2 
\end{pmatrix}$ for some $y_1,y_2\in \mathbb{F}_q$, we have \begin{align*}
 f(t\omega)&= \mu(t)f(\omega)    \\
 &= f(\omega\omega^{-1}t\omega)=f(\omega)\chi(\omega^{-1}t\omega) \implies  0=f(\omega)\big(\mu(t)-\chi(\omega^{-1}t\omega)\big). 
 \end{align*}
So $f(\omega)\neq 0$ then $\mu(t)=\chi(\omega^{-1}t\omega)=\chi_1(y_2)\chi_2(y_1)=:\omega_\chi(t)$ for all $t\in T$, where $\omega_\chi(t):=\chi_1(y_2)\chi_2(y_1)$. So, if there exists a function $f\in V$ with $f(\omega)\neq 0$ then $\mu=\omega_\chi$ on $T$. Conversely, if $\mu=\omega_\chi$ on $T$ then there exists a function $f\in V$ with $f(\omega)\neq 0.$
Subquestion. How could we conclude the proof from these observations? Thanks!
 A: Seems like you've got most of the proof already. You've shown that there exists $f \in V$ such that $f(1) \ne 0$ iff $\chi_1 = \mu_1$ and $\chi_2 = \mu_2$, and there exists $f \in V$ such that $f(\omega) \ne 0$ iff $\chi_1 = \mu_2$ and $\chi_2 = \mu_1$. And you also correctly noted that $G = B \sqcup B \omega B$ which means $f$ is determined by $f(1)$ and $f(\omega)$.
In other words, we have to show that $\operatorname{Ind}_B^G \chi \cong \operatorname{Ind}_B^G \mu$ is equivalent to $V \ne 0$, because any nonzero $f \in V$ would have $f(1) \ne 0$ or $f(\omega) \ne 0$, which would imply $\chi_1, \chi_2 = \mu_1, \mu_2$ in some order.
Note that $\operatorname{Ind}_B^G \chi$ is irreducible iff $\chi_1 \ne \chi_2$. Indeed, $\operatorname{Hom}_G(\operatorname{Ind}_B^G \chi, \operatorname{Ind}_B^G \chi)$ is isomorphic to the vector space of functions $\varphi : G \to \mathbb{C}$ satisfying $\varphi(b_2 g b_1) = \chi(b_2) \varphi(g) \chi(b_1)$. By what we said above, there exists $\varphi$ with $\varphi(1) \ne 0$ but there does not exist $\varphi$ with $\varphi(\omega) \ne 0$, which means this space is one-dimensional. That is, the only $G$-linear maps from $\operatorname{Ind}_B^G \chi$ to itself are homotheties; it has no nontrivial fixed subspaces.
Now we're done by Schur's lemma.
\begin{align*}
\operatorname{Ind}_B^G \chi \cong \operatorname{Ind}_B^G \mu \iff \operatorname{Hom}_G(\operatorname{Ind}_B^G \chi, \operatorname{Ind}_B^G \mu) \ne 0 \iff V \ne 0 \iff \chi_1, \chi_2 = \mu_1, \mu_2
\end{align*}
