Fix a complete theory $T$, a monster model $U\models T$, and a small subset $A\subset U$. We say that an $A$-indiscernible sequence $\mathcal{I}$ is $A$-based when, for any sequences $\mathcal{I}_1,\mathcal{I}_2$ realizing the $\text{EM}$-type of $\mathcal{I}$ over $A$, there exists $c$ such that each $\mathcal{I}_l+(c)$ is $A$-indiscernible. The following is Exercise 2.45 in Pierre Simon's book on NIP theories:
Let $(a_i)_{i<\omega2}$ be indiscernible. Show that $(a_i)_{i<\omega}$ is based on $(a_i)_{\omega\leqslant i<\omega 2}$. Conclude that any indiscernible sequence is based on some set.
I've found that there is a slight generalization of the problem that has been (for me) a bit more convenient to think about:
Suppose $\mathcal{I},\mathcal{J},\mathcal{K}$ are indiscernible sequences such that $\mathcal{I}+\mathcal{K}$ and $\mathcal{J}+\mathcal{K}$ are both indiscernible. Then there exists $c$ such that $\mathcal{I}+(c)$ and $\mathcal{J}+(c)$ are both $\mathcal{K}$-indiscernible.
When we take $\mathcal{K}$ to have order type $\omega$, I think this is precisely equivalent to the first formulation of the question, so I will use the latter formulation throughout the rest of the post. If the order type of $\mathcal{K}=(e_k)_{k\in K}$ has no least element, then I think the desired result is straightforward by compactness; indeed, for any $k\in K$, we know that $\mathcal{I}+(e_k)$ and $\mathcal{J}+(e_k)$ are $(e_l)_{l>k}$-indiscernible. But this approach breaks down if $K$ does have a least element, and it's not intuitively clear to me that the desired result should hold in that case.
For example, suppose $T$ is the theory of dense meet trees, in the language $\{<,\wedge\}$. (See section 2.3.1 of Simon's book for details.) Let $\mathcal{K}=(e_k)_{k\in\omega}$ be any strictly decreasing sequence. By definition of $T$, there are infinitely many open cones of $U$ with center $e_0$; if we take $\mathcal{I}=(a_i)_{i\in\omega}$ and $\mathcal{J}=(b_j)_{j\in\omega}$ to be any strictly decreasing sequences lying in two distinct open cones with center $e_0$, then I believe $\mathcal{I}+\mathcal{K}$ and $\mathcal{J}+\mathcal{K}$ will each be indiscernible. However, by definition of open cones, we have $a_i\wedge b_j=e_0$ for any $i,j\in\omega$; in particular, there does not exist any $c$ such that $\mathcal{I}>c>e_0$ and $\mathcal{J}>c>e_0$, and it is hence impossible to find $c$ such that $\mathcal{I}+(c)$ and $\mathcal{J}+(c)$ are $\mathcal{K}$-indiscernible. Am I making a mistake somewhere in this example?
For the second half of the problem, I believe we only do need the case where $K$ has no least element; indeed, given an indiscernible sequence $\mathcal{I}=(a_i)_{i\in I}$, by compactness we can find an indiscernible sequence of order type $I+\omega^*$ realizing the $\text{EM}$-type of $\mathcal{I}$. After an automorphism moving the first $I$ elements of this sequence to $\mathcal{I}$, we get an indiscernible sequence $\mathcal{K}$ with order type $\omega^*$ such that $\mathcal{I}+\mathcal{K}$ is indiscernible; then, by the discussion above, $\mathcal{I}$ will be $\mathcal{K}$-based. But I'm wondering about how much this argument really does depend on the order type of the sequence in question; any insight would be appreciated!
Your analysis is correct. In the exercise, the indiscernible sequence should be indexed by $\omega+L$, where $L$ is some linear order with no least element.
Here's another perspective on what's going on here: In the context of NIP theories, an indiscernible sequence $I$ is "based on" a set $A$ if and only if $I$ is a Morley sequence over $A$ for a global $A$-invariant type $p$ (one direction Lemma 2.41 in A Guide to NIP Theories, and the other direction is easy: given $I_1$ and $I_2$ realizing the EM-type of $I$ over $A$, take $a$ to be any realization of $p$ over $A$ and $I_1$ and $I_2$). Now if $I+K$ is an indiscernible sequence such that $K$ has no least element, then $I$ is a Morley sequence over $K$ in the global type $\varprojlim K$ (where I use the $\leftarrow$ to indicate that we take the limit type of $K$ in its reverse order!), which is finitely satisfiable in $K$. Thus $I$ is based on $K$. This argument fails when $K$ has a least element.