I want to prove that a sequence of elements in ($Q, <$) is indiscernible if and only if it is either strictly increasing or strictly decreasing or constant.
For the reverse direction: I note that the ($Q, <$) is a model of dense linear order(DLO). Then by quantifier elimination in DLO, the type of a tuple ($x_1, x_2, ..., x_n$) over $\varnothing$ is completely determined by the relative ordering of $x_i$s. So, take a strictly increasing sequence $a_1<a_2 ...,$, if $i_1, ..., i_n$ and $j_1, ...,j_n $ then, type($a_{i_1}, ..., a_{i_n} $/$\varnothing$) = type($a_{j_1}, ..., a_{j_n} $/$\varnothing$) since both the tuples (type($a_{i_1}, ..., a_{i_n} $) and type($a_{j_1}, ..., a_{j_n}$) are strictly increasing. So, the sequence is indiscernible.
I think the same (exact) argument can be used for strictly decreasing and constant sequences.
However, I don't know how to do the forward direction? I have to show that $\phi(a_{i_1}, ..., a_{i_n})$ $\leftrightarrow$ $\phi(a_{j_1}, ..., a_{j_n} )$ holds only when the sequence is strictly increasing/decreasing or constant. Am I supposed to prove this by contradiction?
The forward direction is actually very simple: given a nontrivial indiscernible sequence $\bar a = (a_i)_{i\in I}$, if you pick any $i_1<i_2\in I$, then, since DLO implies that $<$ is a total ordering, you have either $a_{i_1}<a_{i_2}$, $a_{i_1}=a_{i_2}$, or $a_{i_1}>a_{i_2}$, and indiscerniblity trivially implies that in each case, $\bar a$ is strictly increasing, constant or strictly decreasing (respectively).