Throughout, work in $\mathsf{ZFC}$ + large cardinals (let's say a proper class of Woodin limits of Woodins but I'm happy to go higher if that would help).

Let $\mathcal{R}=(\mathbb{R};+)$ be the additive group of real numbers. In $V$, we have $\mathit{Aut}(\mathcal{R})\cong S_\mathbb{R}$ - that is, the automorphism group of $\mathcal{R}$ is abstractly just the group of permutations of any fixed size-continuum set.

Now in $L(\mathbb{R})$ this does not hold due to large cardinals. However, I don't see that this means that $L(\mathbb{R})$ must not contain any copy of $\mathit{Aut}(\mathcal{R})$:

Is there a group $G\in L(\mathbb{R})$ such that, in $V$, $G\cong \mathit{Aut}(\mathcal{R})$?

Unless I'm missing something, $(S_\mathbb{R})^{L(\mathbb{R})}$ is much too small to be a candidate for this: $L(\mathbb{R})$ is certainly missing most permutations of the reals! That said, I see no reason why a copy of $\mathit{Aut}(\mathcal{R})$ couldn't show up in $L(\mathbb{R})$ somewhere "far away" from $\mathcal{R}$ itself.

More generally, is there a structure $\mathcal{S}$ in $L(\mathbb{R})$ whose $V$-automorphism group does not have a copy $\mathcal{G}\in L(\mathbb{R})$? (Note that - here and above - I don't demand that $L(\mathbb{R})$ in any way be able to connect elements of this strange copy $\mathcal{G}$ with $\mathcal{S}$; in particular, $L(\mathbb{R})$ need not have any interesting maps $\mathcal{G}\times\mathcal{S}\rightarrow\mathcal{S}$.)

Throughout, work in $\mathsf{ZFC}$ + large cardinals (let's say a proper class of Woodin limits of Woodins but I'm happy to go higher if that would help).

Let $\mathcal{R}=(\mathbb{R};+)$ be the additive group of real numbers. We have $\mathit{Aut}(\mathcal{R})^{L(\mathbb{R})}\not\cong\mathit{Aut}(\mathcal{R})^V$ due to large cardinals. However, I don't see that this means that $L(\mathbb{R})$ must not contain any copy of $\mathit{Aut}(\mathcal{R})$:

Is there a group $G\in L(\mathbb{R})$ such that, in $V$, $G\cong \mathit{Aut}(\mathcal{R})$?

Unless I'm missing something, the set of all functions from reals to reals in ${L(\mathbb{R})}$ is too small to be a candidate for this. That said, I see no reason why a copy of $\mathit{Aut}(\mathcal{R})$ couldn't show up in $L(\mathbb{R})$ somewhere "far away" from $\mathcal{R}$ itself.

More generally, is there a structure $\mathcal{S}$ in $L(\mathbb{R})$ whose $V$-automorphism group does not have a copy $\mathcal{G}\in L(\mathbb{R})$? (Note that - here and above - I don't demand that $L(\mathbb{R})$ in any way be able to connect elements of this strange copy $\mathcal{G}$ with $\mathcal{S}$; in particular, $L(\mathbb{R})$ need not have any interesting maps $\mathcal{G}\times\mathcal{S}\rightarrow\mathcal{S}$.)