Here's Attempt #2 at asking this question.
Suppose $X$ is a set (not necessarily finite), and let $T$ denote the collection of all totally ordered sets $(A,\leq)$ such that $A \subseteq X.$ Now let $\cong$ denote the unique relation on $T$ such that $(A,\leq_A) \cong (B,\leq_B)$ holds iff they're isomorphic as totally ordered sets.
Question. What is the cardinality of $T /\!\cong$ in terms of the cardinality of $X$?
If $X$ is finite, this is easy: the answer is $|T/\!\cong\!| = |X|+1.$ So its really the infinite case that I'm interested in. For the infinite case, all I know is that $|X|^+ \!\leq |T/\!\cong\!|,$ where $\kappa^+$ denotes the cardinal successor of $\kappa$. This follows more or less from the definition of cardinal successor.