Each and every function $f:X\rightarrow Y$ between two totally ordered sets $(X,\leq_X), (Y,\leq_Y)$ induces a relation $\preceq$ on $X$ by
$$x \preceq x' :\equiv \begin{cases} x &\leq_X x' &\mbox{if } &f(x) = f(x') \\ f(x) &\leq_Y f(x') &\mbox{if } &f(x) \neq f(x') \end{cases} $$
Is this relation $\preceq$ more "natural" than other relations definable over $\leq_X$, $\leq_Y$?
Is $\preceq$ automatically a total order? Or: Which extra conditions do $\leq_X$, $\leq_Y$ have to fulfill such that $\preceq$ is a total order?
Assume that $\leq_X$, $\leq_Y$ are well-orders. Which conditions do they have to fulfill such that $\preceq$ is a well-order again?
Let $\le$ be the lexicographic order on $Y\times X$: $\langle y_0,x_0\rangle\le\langle y_1,x_1\rangle$ iff either $y_0\le_Y y_1$, or $y_0=y_1$ and $x_0\le_X x_1$. Then the map
$$X\to Y\times X:x\mapsto\langle f(x),x\rangle$$
is an order-isomorphism between $\langle X,\preceq\rangle$ and $\langle f^{-1},f^{-1}\,\cap\le\rangle$. (I view $f$ as a subset of $X\times Y$.) Thus, $\preceq$ is always a linear order. If $\le_X$ and $\le_Y$ are well-orders, so is $\le$, and therefore so is $\preceq$.