Embeddings from L-ranks into themselves It is well-known that the existence of a (nontrivial) elementary embedding from a rank $V_\lambda$ (from the von Neumann hiearchy) into itself (i.e. $j:V_\lambda\to V_\lambda$) is a very powerful axiom, as its critical point would be n-huge for all n.
I was wondering what would be the strength of:
(L) There is a nontrivial elementary embedding $j: L_\lambda\to L_\lambda$ (such that $\lambda$ is the supremum of the critical sequence $\{j^n(\kappa)|n\in\mathbb{N}\}$)
I know that the critical point of such an embedding cannot be measurable (as IIRC "there is a measurable" implies that there exists $j:L\to L$).
EDIT: I wrote "as its critical point would be super-n-huge for all n."
This is plain false (it's only n-huge for all n) and I'm sorry if I misinformed anyone.
 A: Over ZFC, the statement "There is a nontrivial elementary embedding $j: L_\lambda\rightarrow L_\lambda$ for some $\lambda$" follows immediately from a measurable. To see this, we argue as follows:


*

*If $j:V\rightarrow M$ is an elementary embedding of $V$ into some inner model $M$, then for each $\alpha$ we have $j\upharpoonright V_\alpha: V_\alpha\rightarrow V_{j(\alpha)}^M$ is also elementary.

*If we set $\lambda=j^\omega(\kappa)$, then $j(\lambda)=\lambda$.

*Since $L$ is definable and absolute between inner models, we also have that $j\upharpoonright L: L\rightarrow L$ is elementary (and with the same critical point).
Combining these facts, we get that $j$ restricts to an elementary embedding $L_\lambda\rightarrow L_\lambda$.
More precisely, I suspect that over ZFC it's equivalent to $0^\sharp$. (You certainly get a sequence of $L$-indiscernibles out of it!)
But your last sentence suggests to me that you're interested in this principle over ZFC+V=L. In that case, the answer is that it is inconsistent. This is via the usual construction of a measure on $\kappa= crit(j)$ from an embedding: $$\mathcal{U}_j=\{x\subseteq\kappa: \kappa\in j(x)\}.$$

Let me elaborate on the construction of $\mathcal{U}_j$. The more general context is the following. Suppose $M, N$ are inner models and $j:M\rightarrow N$ is nontrivial elementary with critical point $\kappa$. Then we can form the set $$\mathcal{U}^M_j=\{x\in\mathcal{P}^M(\kappa): \kappa\in j(x)\}.$$ Obviously there's no reason to believe $\mathcal{U}^M_j\in M;$ e.g. if $M=L$, then it never will be. But regardless of what $M$ is, $\mathcal{U}^M_j$ will be a $\kappa$-complete nonprincipal ultrafilter "relative to $M$," in the following sense:


*

*No finite set is in $\mathcal{U}_j^M$.

*If $x\in\mathcal{P}^M(\kappa)$, then $x\in\mathcal{U}^M_j$ or $\kappa\setminus x\in\mathcal{U}_j^M$.

*If $x\subset y$ are subsets of $\kappa$ in $M$ and $x\in\mathcal{U}^M_j$, then $y\in\mathcal{U}_j^M$.

*If $\langle x_\eta\rangle_{\eta<\mu}$ is a sequence in $M$ of subsets of $\kappa$ (that is, the whole sequence is in $M$, not just each individual term) with $\mu<\kappa$, and $x_\eta\in\mathcal{U}_j^M$ for all $\eta$, then $\bigcap_{\eta<\mu} x_\eta\in\mathcal{U}^M_j$.
The first three points are immediate. The last one's a bit subtle. The idea is that since $\mu<\kappa$, $j$ will commute with basic set operations: in particular, we'll have $$(*)\quad j(\bigcap_{\eta<\mu} x_\eta)=\bigcap_{\eta<\mu} (j(x_\eta)).$$ The right hand side clearly contains $\kappa$, so if we can show this we'll have $\bigcap_{\eta<\mu} x_\eta\in \mathcal{U}^M_j$.
To see why $(*)$ is nontrivial, the point is that if a sequence has length $\kappa$ then it "stretches" when we hit it with $j$ - it gains "nonstandard" terms. To understand this, think about the following: if we let $x_\eta=\kappa\setminus\eta$, then each $x_\eta$ is in $\mathcal{U}^M_j$ but $\bigcap_{\eta<\kappa} x_\eta=\emptyset\not\in\mathcal{U}^M_j$.
A: The existence of $0^{\#}$ suffices to produce an elementary embedding $j:L_\lambda\to L_\lambda$, where $\lambda$ is the $\omega$-th Silver indiscernible $i_\omega$. The embedding $j$ sends each of the previous Silver indiscernibles $i_n$ to the next one, $i_{n+1}$, and it is automatically defined on all other elements of $L_{i_\omega}$ because it has to commute with the definable Skolem functions.
(If I remember correctly, the existence of such an elementary embedding is in fact equivalent to the existence of $0^{\#}$.) 
