Characterizing elementary embeddings of $L$ and $L_\alpha$ under 0# 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.)
 A: The answer to question 1 is no.  Let $n$ be a finite ordinal, and consider the structure $M$ with universe $L_{i_n}$, with constant symbols for the smaller Silver indiscernibles $i_0,\dots,i_{n-1}$ as well as symbols for the membership relation $\in$ and the usual, $L$-definable Skolem functions.  This structure $M$ is constructible.  (This is where it's essential that $n$ is finite.)  So the Skolem hull, the smallest elementary substructure $N$ of $M$, is constructibly countable.  But $M$ itself is very large in the sense of $L$, since Silver indiscernibles like $i_n$ are constructibly inaccessible (and much more).  Therefore $N$ is not all of $M$.
A: The answer to Q2 is 'No'.  Suppose $j:L\rightarrow L$ is a non-trivial elementary embedding. We use the following fact:
$\bullet$ $cp(j)$  (the first ordinal moved by $j$) is always a Silver indiscernible.
Now let $I$ be the class of Silver indiscernibles, and $\delta \in I$ but $j(\delta)\notin I$ for a contradiction.  Let $H$ be the Skolem hull in $L$ of $j(\delta)\cup j$''$I\backslash (\delta +1)$. $H$ is isomorphic to $L$. If $j(\delta)\notin H$ but $\pi:H \rightarrow L$ is the transitive collapse, then $\pi^{-1}:L\rightarrow L$ is non-trivial with critical point $j(\delta)$. Hence, by the Fact,  $j(\delta)$ must be in $H$. Then we see that for some $\vec \xi <j(\delta)$ some $\overrightarrow{j(\zeta)} > j(\delta)$ with $\vec \zeta \in I\backslash (\delta +1)$ that
$L \models $ ''$\exists \vec \xi < j(\delta)( j(\delta) = t(\vec \xi  ,\overrightarrow{j(\zeta)}))$''.
for some term $t$. But then:
$L \models $ ''$\exists \vec \xi < \delta(  \delta = t(\vec \xi, \overrightarrow{\zeta}))$''
is a definition of the indiscernible $\delta$ from larger indiscernibles and smaller ordinals, which is impossible.  (This works for the variant of the question, taking embeddings between sets, if $\alpha, \beta$ are limit ordinals.)
