Let $T$ be a stable theory. Every Indiscernible sequences is a Indiscernible sets. (Where Indiscernible set is a Indiscernible sequence where every permutation on the order keeps it Indiscernible sequence.)
So what i tried to do is to show that given an Indiscernible sequence $\cal I=(a_i)_{i\in I}$ To show that for all permutations $\sigma$ of the form $(i,j)$ that $tp(a_{i_1},.. a_{i_n})=tp(a_{\sigma (i_1)},..,a_{\sigma (i_n)})$ and since the permutations $(i,j)$ generate the group of permutations on $I$ then it is enough. To do that, I think that it have something to do with the order propery, since we know that if a formula is stable then it doesn't have the order property. But I'm not sure how to use it
Stronger, the permutations of $\{1,\dots,n\}$ are generated by transpositions of the form $(i,i+1)$. So there is some $n$ and $i<n$ such that $\text{tp}(a_1,\dots,a_i,a_{i+1},\dots,a_n) \neq \text{tp}(a_1,\dots,a_{i+1},a_i,\dots,a_n)$. Then they disagree on a formula $\varphi(x_1,\dots,x_i,x_{i+1},\dots,x_n)$.
Now if $\varphi$ only had two free variables ($i = 1$ and $n = 2$), then it would have the order property (do you see why?). You just have to figure out what to do with the extra elements $a_1,\dots,a_{i-1}$ and $a_{i+2},\dots,a_n$. Hint: Push them far away, and add them all as constant elements of the tuples witnessing the order property.