No, such embeddings can consistently exist in $L$. In the following, $\omega_1$ and $\omega_2$ will denote these cardinals as computed in $V$.
Let us assume that $0^\sharp$ exists in $V$. There is an increasing sequence $\langle \xi_n\mid n<\omega\rangle\in V$ of order indiscernibles over the structure $(L_{\omega_2};\in, \alpha\mid\alpha<\omega_1)$, for example take the first $\omega$-many Silver indiscernibles $\geq\omega_1$. In $L$, we can define the tree $T$ which attempts to find such a sequence. As $T$ is illfounded in $V$, $T$ is illfounded in $L$, i.e. there is such a sequence $\langle \xi_n\mid n<\omega\rangle$ in $L$. Now let $X=\mathrm{Hull}^{L_{\omega_2}}(\omega_1\cup\{\xi_n\mid n<\omega\})$ and let $\pi\colon X\rightarrow L_\gamma$ be the transitive collapse of $X$. Then $\omega_1<\gamma$. Let us write $\bar \xi_n=\pi(\xi_n)$. We now have
- $L_\gamma=\mathrm{Hull}^{L_\gamma}(\omega_1\cup\{\bar \xi_n\mid n<\omega\})$ and
- the $\langle \bar\xi_n\mid n<\omega\rangle$ are order indiscernible over $(L_\gamma;\in,\alpha\mid\alpha<\omega_1)$.
This allows us to build elementary embeddings $j\colon L_\gamma\rightarrow L_\gamma$ similar to how one builds elementary embeddings $j\colon L\rightarrow L$ from Silver indiscernibles. Here is the construction:
Let $i\colon\omega\rightarrow\omega$ be any strictly increasing function. We can define an elementary $j\colon L_\gamma\rightarrow L_\gamma$, $j\in L$, by
$j(\tau^{(L_\gamma;\in, \alpha\mid\alpha<\omega_1)}(\bar\xi_0,\dots,\bar\xi_n))=\tau^{(L_\gamma;\in,\alpha\mid\alpha<\omega_1)}(\bar\xi_{i(0)},\dots,\bar\xi_{i(n)})$ for any term $\tau(x_0,\dots, x_n)$ in the language with the $\in$-relation and a constant for every ordinal below $\omega_1$. It is not difficult to check that $j$ is well-defined, elementary and, in case $i\neq\mathrm{id}_{\omega}$, even non-trivial.