Suppose 0# exists.

It is clear that every order preserving map from the indiscernibles to the indiscernibles gives an elementary embedding from $L$ to $L$. Furthermore, following lemmas 18.7 and 18.8 of Jech, if $\alpha$ is an  *infinite  infinite limit* ordinal, an increasing map from alpha to beta gives an elementary embedding from $L\_{i\_\alpha}$ to $L\_{i\_\beta}$, where $i\_\alpha$ is the $\alpha$-th indiscernible. This is because $L\_{i\_\alpha}$ equals the Skolem hull in itself of the first $\alpha$ indiscernibles. However, I am not clear on the following points.

1) Is it the case that for a *finite successor* ordinal, n, $L\_{i\_n}$ is necessarily equal to the Skolem hull in $L\_{i\_n}$ of the first n indiscernibles? Jech only proves this result for infinite ordinals. 

2) Is it possible that there could be an elementary embedding from $L$ to $L$, or from $L\_{i\_\alpha}$ to $L\_{i\_\beta}$ ($\alpha, \beta$ may be finite or infinite), that does not always map indiscernibles to indiscernibles? 
This sounds weird, but I'm not convinced it's impossible. As far as I know, there's no formula in $L$ that defines "$\alpha$ is a Silver indiscernible." (In fact there is no such formula -- see Andreas Blass's comment below.)