Let $X$ be a $0$-type definable set and $E$ a $0$-type definable equivalence relation.
One can prove (see for example proposition 16.3 in Enrique Casanovas' stability notes http://www.ub.edu/modeltheory/documentos/stability.pdf) that if we have an indiscernible sequence $([a_i]_E)_{i\in I}$ of hyperimaginaries. There exist a sequence $(b_i)_{i\in I}$ of representatives that is also indiscernible.
Are there any well-known results if we substitute indiscernibility by total indiscernibility? (At least for real parameters?).
This is not true in general, even for imaginaries.
Consider $M = (\mathbb{Q};<,E)$, where $E$ is an equivalence relation with infinitely many classes such that every $E$-class is dense with respect to the order $<$. This is the Fraïssé limit of the class of all finite linear orders equipped with an arbitrary equivalence relation.
Any sequence of distinct $E$-imaginaries is totally indiscernible, but no sequence of representatives is totally indiscernible, due to the ordering.
To prove the claim about total indisceribles in $\mathbb{Q}/E$, it suffices to show that for any distinct $a_1,\dots,a_n\in\mathbb{Q}/E$ and distinct $b_1,\dots,b_n\in\mathbb{Q}/E$, there is an automorphism of $M$ mapping each $a_i$ to $b_i$ (i.e., the induced structure on $\mathbb{Q}/E$ is that of an infinite set with equality). Pick a representative $a_i'$ for each $a_i$ such that $a_1'<a_2'<\dots< a_n'$. This is possible since each $E$-class is dense. Similarly, pick a representative $b_i'$ for each $b_i$ such that $b_1'<b_2'<\dots< b_n'$. Then the tuples $(a_1',\dots,a_n')$ and $(b_1',\dots,b_n')$ have the same quantifier-free type in $M$ (the type of an increasing sequence of pairwise non-$E$-equivalent elements), so by ultrahomogeneity of $M$ there is an automorphism mapping each $a_i'$ to $b_i'$ and hence each $a_i$ to $b_i$.
This example shows that an infinite quotient of an unstable set can be stable. It seems unlikely to me that we could prove any general theorem on pulling back total indiscernibles in $X/E$ to total indiscernibles in $X$, except in the trivial case that $X$ itself is stable (in which case all indiscernibles in $X$ are total indiscernibles).