I have read somewhere (unfortunately I cannot locate the exact text anymore) that there exists an equality in distribution between the order statistics of any continuous distribution and the uniform distribution.
In other words, let $X_1,\ldots,X_n$ be identically distributed random variables, where $X_{1:n},\ldots,X_{n:n}$ are their order statistics so that $X_{1:n}<\ldots<X_{n:n}$. Then, there exists an equality in distribution
\begin{equation*} \frac{X_{i:n}}{X_{n:n}}\overset{d}{=} Y_{i:n-1}, \quad \forall 1\leq i\leq n-1 \end{equation*} where $Y_i\sim \text{Uniform}(0,1)$. In other words, the ratio $\frac{X_{i:n}}{X_{n:n}}$ has supposedly the same distribution as the $i\text{th}$ order statistic of the $n-1$ independent and uniformly distributed random variates $U_1,\ldots,U_{n-1}$. Is this true?
Can we then say that $\frac{X_{1:n}}{X_{n:n}},\ldots,\frac{X_{n-1:n}}{X_{n:n}}$ are i.i.d. and distributed as $\sim \text{Uniform}(0,1)$?
$\textbf{Edit:}$
I managed to track down the text I was thinking of. It is in the book "A First Course in Order Statistics" by Barry C. Arnold , N. Balakrishnan and H. N. Nagaraja. Specifically, I am thinking of equations (2.4.1) and (2.4.2) in Chapter 2, pp. 21-22.
Please have a look on Google Books (I do not want to copy-paste the whole thing due to copyright): 2.4. Some Properties of Order Statistics