Assume a random sample $X_1,X_2, \dots , X_n$ following the Gamma distribution with pdf $$ f(x) = \frac{b^a}{\Gamma (a)} x^{a-1} e^{-bx}, \, x>0 $$
($a$ known, $b$ unknown).
The distribution of $T= \sum\limits_{i=1}^n x_i$ follows the Gamma distribution with parameters $na$ and $b$. Also, if $F_T(t)$ is the cdf of $T$, we have that the sample mean $\bar{X} = \frac{1}{n} T$ has cdf $F_T(nt)$.
Having said that, we must show that:
$Y = b\bar{X}$ cdf does not depend on $b$.
$$ F_Y(y) = P(Y \leq y) = P\left(T \leq \frac{ny}{b}\right)=F_T\left( \frac{ny}{b} \right) $$ Can the cdf of $Y$ be explicitly calculated to show that it does not depend on $b$?
The density function of $T$ is $\phi(y)= b^{na} y^{na-1}e^{-by} (\Gamma (na))^{-1}$ and (by differentiation ) the density of $Y$ is $\frac n b \phi (\frac {ny} b)$. Can you see that $b$ cancels out in this expression?