Quantile(X * k) = Quantile(X) * k? I just wanted to verify that this is true.
Say the .95th quantile of a random variable, $X$, is $q$. Then the .95th quantile of $\frac{X}{k}$ is $\frac{q}{k}$ where $k$ is a constant.
I am assuming this property holds.
$$
P(X*k > x) = P\left(X > x * \frac{1}{k}\right)
$$

Short answer: yes, if $k>0$ because in that case, the two statements under the probability sign are logically equivalent, so they are really just two slightly different ways to describe the exact same underlying event.
Longer answer (assuming knowledge of some basic measure-theoretic probability definitions): Remember that probabilities are really functions over certain (i.e. measurable) sets. Thus, if $X:\Omega\to\mathbb R$ is a random variable on the measure space $\Omega$, a quantity denoted $P(X\times k > x)$ is really just syntactic sugar for a more pedantic statement like $P(\{\omega\in \Omega: X(\omega) \times k > x\})$. But now, note that if $k>0$,
$$P(X\times k > x) \equiv P(\{\omega\in \Omega: X(\omega) \times k > x\}) = P(\{\omega\in \Omega: X(\omega) > x / k\}) \equiv P(X>x/k)$$
The two outer equalities are definitions while the middle one follows by basic algebra showing that the conditions defining the respective sets are the same.
