For an arbitrary sequence $X$ of $n$ distinct real numbers, can we uniquely and exhaustively define a set of infinitely many pairs of the form: $[f_{j},$ order$(f_{j}(x))]$, where $f_{j}$ is a real-valued function that returns a sequence of $n$ real numbers, and is neither piecewise nor constant, and order$(.)$ returns the ascending rank order of the elements of $f_{j}(x)$?
For example, suppose $n = 4$ and $X = (.1,0,.2,44)$, and $f_{1}(x) = x^2 = (.01,0,.4,1936)$, $f_{2}(x) = (x - 2)^2 = (3.61, 4, 3.24, 1764)$, and so on. Then its pairs would be $[f_{1}, (2,1,3,4)], [f_{2}, (2,3,1,4)]$, and so on. If we define such pairs for $X$ and all possible $f_{j}$, must there be at least one other sequence $Y$ of $n$ distinct real numbers that, when plugged into all the same functions, gives all the same answers in terms of rank order of elements? Or will all $Y$ fail for at least one $f_{j}$?
Here is a countable set $\cal F$ of functions that serve to uniquely identify every sequence of $n\ge 3$ real numbers in the manner you propose. Define $f_r$ by $$f_r(x):= (x-r)^2,$$ and let ${\cal F}$ be the collection of all such $f_r$ as $r$ ranges over the rational numbers. Since $$f_r(x) - f_r(x') = (x-x')(x+x'-2r),$$ for given slots $i$ and $j$ in the output vector you can deduce the value of $x_i+x_j$ by checking for each $r$ whether $f_r(x_i)$ ranks higher or lower than $f_r(x_j)$. (Specifically, there will be a flip in rank order when $2r$ passes the threshold $x_i+x_j$.)
So if $X=(x_1,\ldots,x_n)$ is a sequence of values , you can inspect the output vectors to determine the sum $x_i+x_j$ for every $i\ne j$. If $n\ge3$, this knowledge is sufficient to deduce the individual values of each $x_i$ and $x_j$.