Does any known large cardinal axiom implies that there exists an elementary embedding $j:L(\mathbb{R})\to L(\mathbb{R})$ ?
Since the Axiom of Determinacy is consistent with $V=L(\mathbb{R})$ I first didn't think it was consistent with the existence of a such embedding but the usual proof that there is no embedding $j:V\to V$ relies on the Axiom of Choice which is provably false under AD so I'm not really sure.
However now I remember seeing papers about how $I0$ was an "AD-like" axiom (with $L(V_{\lambda+1})$ instead of $L(\mathbb{R})$), and $I0$ is precisely about the existence of an elementary embedding $j:L(V_{\lambda+1})\to L(V_{\lambda+1})$, so $AD^{L(\mathbb{R})}$ implying (or being implied by) the existence of an embedding $j:L(\mathbb{R})\to L(\mathbb{R})$ doesn't seem so implausible anymore.
I might be missing something, but I think a measurable already does this. From a measurable of course we get an elementary embedding $j:V\rightarrow M$ with critical point $\kappa$. $\kappa$ is uncountable, so we have $\mathbb{R}^V=\mathbb{R}^M$; since $L$ is absolute, we further have $L(\mathbb{R})^V=L(\mathbb{R})^M$. Finally, since $L(\mathbb{R})$ is definable we know that $j$ restricts to an elementary embedding $L(\mathbb{R})\rightarrow L(\mathbb{R})$, and that this embedding is nontrivial since $\kappa\in L(\mathbb{R})$, being an ordinal.
I do not, off the top of my head, see how to get such an embedding in ZF+ V=L$(\mathbb{R})$, and I suspect that none can exist (especially if we throw AD into the mix); however, since choice fails Kunen's argument doesn't go through here, so I don't see the disproof.