$B(T)$ and $B^a(T^a)$ are same distribution? ($T$ and $T^a$ are same distribution)

Question

Let $\{B_1(t):t\geq0\}$ be a one dimensional Brownian motion. Then, $\{B^a_1(t):t\geq0\}$ is also a Brownian motion where $B^a_1(t):=a^{-1}B_1(a^2t)$ for $a\in\mathbb{R}$.

Let $\{B_2(t):t\geq0\}$ be a one dimensional Brownian motion independent of $\{B_1(t):t\geq0\}$. For $b\in\mathbb{R}$, $T:=\inf\{t\geq0:B_2(t)=b\}$ and $T^a:=\inf\{t\geq0:B^a_2(t)=b\}$ have same distribution where $B^a_2(t):=a^{-1}B_2(a^2t)$ for $a\in\mathbb{R}$.

I already know these facts is right.

Question

$B_1(T)$ and $B^a_1(T^a)$ have same distribution?

Thank you for your cooperation.

Answer 1

$B^a(T^a)=B(T)=b$ by definition of your stopping time (note that $T$ and $T^a$ are a.s. finite), so yes.