Constructing a bijection from A to A where |A| is finite satisfying ordering conditions. I'm given two total orderings, set $(A,<)$, and $(A,<')$, where $A$ has a finite cardinality and am trying to prove/disprove that there exists a unique bijection $f:A \rightarrow A$ such that $x < y$ implies $f(x)<'f(y)$ for all $x,y \in A$. 
I've been attempting to construct a function $f$ that satisfies these conditions, but am not sure where to start. The two different orderings are confusing me as I'm not sure how one is to have both $x,y$ and $f(x), f(y)$ ordered differently based on a single function $f$.
Any guidance on where to begin or what to consider would be greatly appreciated.
 A: This is a situation where I think a slightly more general question is actually going to be easier to think about, namely:

Suppose $(A,<)$ and $(B,\triangleleft)$ are two finite linear orders with $\vert A\vert=\vert B\vert$. Show that there is a unique $f:A\rightarrow B$ such that $$x<y\iff f(x)\triangleleft f(y).$$ 

This is a bit easier to think about in my opinion since we've freed up our mental picture by considering two different sets. It's not a substantive difference, but I suspect it will make things a bit more concrete.
Now here's my hint - which a priori is just for existence, but will quickly lead to uniqueness as well:

Where should we map the "first" (in the sense of $<$) element of $A$? What about the "second" element of $A$? Etc.

There are a couple assumptions hidden in this hint which you need to tease out and prove of course; for example, why do we know that $A$ even has a "first" element at all? This is where finiteness will come in, both in the existence and uniqueness steps. (To see why we definitely need finiteness in both pieces, show that existence fails if we consider the naturals and the negative naturals with the usual orderings, and uniqueness can fail if we take both $A$ and $B$ to be the naturals with the usual ordering.)
A: A total order $<$ over a finite set $A$ is always a well-order, because every nonempty subset of $A$ is finite and hence has a minimum.
Hence $(A,<)$ is isomorphic to $(n,\in)$ through a unique order isomorphism $f\colon A \to n$, beign $n$ the unique natural number such that $n = |A|$.
Analogously, $(A,<')$ is isomorphic to $(n,\in)$ through a unique order isomorphism $f'\colon A \to n$.
Hence $(f')^{-1} \circ f$ is the isomorphism $(A,<) \to (A,<')$ that you where looking for. Moreover it is the unique one.
