If $\pi$ is a normal number, would that imply that $\tau =2\pi $ is also a normal number? If so, why? Something tells me that it should be, but I have no idea how to prove it. If all digits of $\pi$ were either $0$, $1$, $2$, $3$ or $4$, the proof is self-evident. Obviously, every decimal digit appears in $\pi$, so that's where that self-evident proof will fall apart.
Yes, a rational multiple of any normal number (with respect to a base $b$, such as $b = 10$) is also normal (with respect to the same base).
The number $\tau = 2 \pi$ is a rational multiple of $\pi$, and visa versa. So, $\tau$ is normal if and only if $\pi$ is normal.