Does $\mathit{Aut}(\mathbb{R};+)$ have a copy in $L(\mathbb{R})$ granting large cardinals? 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 $\mathcal{G}\times\mathcal{S}\rightarrow\mathcal{S}$.)
 A: Assuming also CH, the answer to the more general question is yes,
there is a structure in $L(\mathbb{R})$ (in fact, just the set $\mathbb{R}$, with no additional structure), whose automorphism group in $V$ does not have an isomorphic copy in $L(\mathbb{R})$. (I haven't really thought about the original question, i.e. $(\mathbb{R},+)$.)
Assume ZFC + CH and $\mathbb{R}^{\#}$ exists
(i.e. there is an elementary embedding $j:L(\mathbb{R})\to L(\mathbb{R})$).
Consider the set $\mathbb{R}$ of reals
with no additional structure. So automorphisms
are just bijections $\pi:\mathbb{R}\to\mathbb{R}$. Let $G$ be the group of such bijections $\pi$ (under composition). I claim that $L(\mathbb{R})$ has no group which is (in $V$) isomorphic to $G$.
The proof will use the existence of $\mathbb{R}^{\#}$ to decompose the supposed copy $G'$ of $G$, with $G'\in L(\mathbb{R})$,
into the direct limit of elementary substructures $G'_s$, each of which have cardinality $\leq$ the continuum (which is $\aleph_1$ under the CH hypothesis), and such that there are only countably many isomorphism types of $G'_s$ over some base substructure $G'_{s_0}$, modulo isomorphisms which fix $G'_{s_0}$ pointwise. Also, $G'_{s_0}$ will be rich enough that, transporting this over to $G$ and corresponding $G_{s_0}$ and $G_s$, we can construct a bijection (an element of $G$) which couldn't appear in any of the $G_s$'s, via a diagonalization.
For $x,y\in\mathbb{R}$
with $x\neq y$ let $\pi_{xy}:\mathbb{R}\to\mathbb{R}$ be the bijection
such that $\pi_{xy}(x)=y$ and $\pi_{xy}(y)=x$ and $\pi_{xy}(z)=z$ for all other $z$.
Lemma 1: Let $x\neq y$. Then the commutator $C_{\pi_{xy}}$ of $\pi_{xy}$ is the set of all $\sigma\in G$ such that $\sigma(x)=y$ and $\sigma(y)=x$.
Lemma 2: Let $x\neq y$. Let $\pi\in G$
be such that the commutator $C_\pi$ of $\pi$ is such that $C_{\pi_{xy}}\subsetneq C_\pi$. Then $\pi=\mathrm{id}$, so $C_\pi=G$.
So the collection of $\pi_{xy}$'s
is definable over $G$ as those
group elements $\pi$ whose commutator
is maximal for being a proper subset of $G$.
In particular, any $G'\preccurlyeq G$
will contain infinitely many $\pi_{xy}$'s.
Lemma 2.5: Let $x,y,z$ be pairwise distinct. Then
$\pi_{xy}\pi_{yz}\pi_{xy}=\pi_{xz}$.
Lemma 2.6:
Let $x\neq y$ and $x'\neq y'$,
such that $\pi_{xy}\neq \pi_{x'y'}$.
Then $\{x,y\}\cap\{x',y'\}=\emptyset$
iff $\pi_{xy}\pi_{x'y'}$ has order 2.
And $\{x,y\}\cap\{x',y'\}\neq\emptyset$
iff $\pi_{xy}\pi_{x'y'}$ has order 3.
Write $\mathscr{I}$ for the class of Silver indiscernibles for $L(\mathbb{R})$.
Let $s\in[\mathscr{I}]^{<\omega}$ with $s\neq\emptyset$. Then write $H_s=\mathrm{Hull}_{\Sigma_1}^{L_{\max(s)}(\mathbb{R})}(\mathbb{R}\cup\{s^-\})$
where $s^-=s\backslash\{\max(s)\})$.
Lemma 3: $H_s$ is countably closed (in $V$).
Proof: Let $\left<x_i\right>_{i<\omega}\subseteq H_s$. Let $\left<\varphi_i,r_i\right>_{i<\omega}$
be such that $\varphi_i$ is a $\Sigma_1$ formula and $r_i\in\mathbb{R}$ and $x_i$ is the unique $x\in L_{\max(s)}(\mathbb{R})$ such that $L_{\max(s)}(\mathbb{R})\models\varphi_i(x,r_i,s^-)$.
Let $r=(\oplus_{i<\omega}r_i,\oplus_{i<\omega}\varphi_i)\in\mathbb{R}$. Then using the parameter $r$,
working in $L(\mathbb{R})$
we can identify $\left<x_i\right>_{i<\omega}$. In fact,
because $\max(s)$ is inaccessible in $L(\mathbb{R})$, there must be some $\gamma<\max(s)$ such that $L_{\gamma}(\mathbb{R})\models\varphi_i(x,r_i,s^-)$
for each $i<\omega$. But then it easily follows that there is a $\Sigma_1$ formula $\psi$ such that $\left<x_i\right>_{i<\omega}$ is the unique $y\in L_{\max(s)}(\mathbb{R})$
such that $L_{\max(s)}(\mathbb{R})\models\psi(y,r,s^-)$,
so $\left<x_i\right>_{i<\omega}\in H_s$,
as desired.
Now suppose that $G'\in L(\mathbb{R})$
is a group isomorphic (in $V$)
to $G$.
So $G'\in H_{s_0}$
for some $s_0\in[\mathscr{I}]^{<\omega}$
(but this doesn't say that $G'\subseteq H_{s_0}$; in fact it isn't,
because $G$ has cardinality $2^{2^{\aleph_0}}$, but $H_{s_0}$
has cardinality $2^{\aleph_0}$).
Let $G'_{s_0}=G'\cap H_{s_0}$.
We have $H_{s_0}\preccurlyeq_{\Sigma_1}^{L_{\max(s_0)}(\mathbb{R})}$,
and since $G'\in H_{s_0}$,
therefore $G'_{s_0}\preccurlyeq G'$
(fully elementary).
Say a group element $g\in G$
is \emph{good}, if $C_g\subsetneq G$
and there is no $h\in G$ such that
$C_g\subsetneq C_h\subsetneq G$
(so equivalently, $g=\pi_{xy}$
for some $x\neq y$).
Say a group element $g\in G'$ is \emph{good} under the same definition
(but we can't characterize this one as being a $\pi_{xy}$;
it's important that the property
is definable just from the group structure). Given $g,h\in G$ which are both good, say that the pair $(g,h)$ is  \emph{independent} if $gh$ has order 2 (cf. Lemma 2.6).
Lemma 4: (i) $G'_{s_0}$ contains uncountably many good elements. In fact,
(ii) there is an uncountable set $A\subseteq G'_{s_0}$ consisting of pairwise
independent good elements.
Proof: Part (i): $L_{s_0}(\mathbb{R})\models$``there are
uncountably many good group elements of $G'$'',
and because $H_{s_0}$ is countably closed
(in $V$), part (i) follows.
Part (ii): Otherwise, there would
be some countable set $A'$ of good elements such that for all good $g\in G'_{s_0}\backslash A'$, there is some $h\in A$
such that $(g,h)$ is dependent.
Let $f:G\to G'$ be an isomorphism.
Let $A=f^{-1}``A'$. Then
we can fix a countable set $C\subseteq\mathbb{R}$ such that for all $g\in A$, we have $g=\pi_{xy}$ for some $x,y\in C$. Let $B'\subseteq G_{s_0}'$ be an uncountable set of good elements
such that $B'\cap A'=\emptyset$.
Let $B=f^{-1}``B'$.
Then we can find an uncountable $\bar{B}\subseteq B$ and an $x\in C$
such that for each $\pi\in\bar{B}$,
there is $y\in\mathbb{R}$ such that
$\pi=\pi_{xy}$. But then taking $y_1,y_2$ such, by Lemma 2.5, we have $\pi_{y_1y_2}=\pi_{y_1x}\pi_{xy_2}\pi_{y_1x}$.
So these generate an uncountable
set of pairwise independent $\pi$'s. Pulling these back under the isomorphism $f$, part (ii) is proven.
Now if $s,t\in\mathscr{I}^{<\omega}$
with $|s|=|t|$ then $H_s\equiv H_t$,
and the isomorhphism fixes $\mathbb{R}$.
Let $\kappa$ be the least limit Silver indiscernible with $\max(s)<\kappa$.
Then $L_\kappa(\mathbb{R})=\bigcup_{s\in[\mathscr{I}\cap\kappa]^{<\omega}}H_s$,
and therefore letting $\mathscr{J}$ be the set of all $s\in[\mathscr{I}\cap\kappa]^{<\omega}$ such that $s_0\subseteq s$, then $G'=\bigcup_{s\in\mathscr{J}}G'_s$.
Also, if $s,t\in\mathscr{J}$ with $s\subseteq t$ then $G'_{s_0}\preccurlyeq G'_s\preccurlyeq G'_t\preccurlyeq G'$,
each $G'_s$ has cardinality continuum
(by Lemma 4
and since $H_s$ has cardinality continuum and by CH).
Lemma 5: Let $s,s'\in\mathscr{J}^{<\omega}$ be such that $|s|=|s'|$
and the $s_0$ sits in ordertype within $s$
just as $s_0$ sits in $s'$. Then $G'_s\cong G'_{s'}$
via an isomorphism which fixes $G'_{s_0}$.
(Here by ``sits in... just as...'' I mean
that if $s=\{s^0,s^1,s^2,\ldots,s^k\}$
with $s^0<s^1<\ldots<s^k$,
and $s'=\{(s')^0,(s')^1,\ldots,(s')^k\}$
increasing likewise,
then for $i\in[0,k]$,
we have $s^i\in s_0$ iff $(s')^i\in s_0$.)
Proof of Lemma 5: Easy consequence of Silver indiscernibility.
Thus, there are only countably many isomorphism types for the subgroups $G'_s$ over $G'_{s_0}$.
For $g\in G'$, let $E_g$
be the set of equations satisfied by
elements in $G'_{s_0}\cup\{g\}$.
(E.g. if $h_1,h_2\in G'_{s_0}$,
we might have the equation ``$h_1gh_2=\mathrm{id}$'' in $E_g$.)
Lemma 6: $\{E_g\bigm|g\in G'\}$
has cardinality at most continuum $=\aleph_1$
(in $V$).
Proof: If $s\in\mathscr{J}$ then $\{E_g\bigm|g\in G'_s\}$ has cardinality at most continuum, since $G'_s$ has at most cardinality continuum. But if $s,s'$
are such that $G'_s$ and $G'_{s'}$
are isomorphic via an isormorphism which fixes $G'_{s_0}$,
then clearly $\{E_g\bigm|g\in G'_s\}=\{E_g\bigm|g\in G'_{s'}\}$. So
by Lemma 5, we get at most continuum
many such $E_g$'s in total.
The structure established for $G'$
now shifts over to $G$ under the isomorphism $f:G'\to G$. Write $G_{s_0}$, $G_s$ etc for the pointwise images of $G'_{s_0}$, $G'_s$ etc.
The rest of the proof is a direct diagonalization, constructing some
$h\in G$ such that $E_h$ is not
one of the $E_g$'s above, which is a contradiction. The construction is just algebraic.
Let $\left<r_\alpha\right>_{\alpha<\aleph_1}$ be an enumeration of $\mathbb{R}$
in ordertype $\aleph_1$. We define by recursion on $\beta<\aleph_1$,
some ordinals $\alpha_\beta<\aleph_1$,
and
$h\upharpoonright\left<r_\alpha\right>_{\alpha<\alpha_\beta}$, with $h``\left\{r_\alpha\right\}_{\alpha<\alpha_\beta}=\left\{r_\alpha\right\}_{\alpha<\alpha_\beta}$. Let $\left<g_\beta\right>_{\beta<\aleph_1}\subseteq G$ be such that all possible $E_g$'s for $g\in G$ are produced by the $E_{g_\beta}$'s (by the lemmas, there are only continuum = $\aleph_1$-many such $E_g$'s).
Suppose we have $\beta<\aleph_1$
and we have defined $\alpha_\beta$
and $h\upharpoonright\{r_\alpha\}_{\alpha<\alpha_\beta}$, as above.
Consider $g_\beta$ and $E_{g_\beta}$.
We want to now make sure that $E_h$
will not be $E_{g_\beta}$. We know
that $G_{s_0}$
has an uncountable subset of pairwise independent good elements. So let $\gamma<\delta<\aleph_1$ be such that $\alpha_\beta\leq\gamma,\delta$
and $\pi_{r_\gamma r_\delta}\in G_{s_0}$.
We can now decide freely whether or not $h\pi_{r_\gamma r_\delta}=\pi_{r_\gamma r_\delta}h$. If $g_\beta \pi_{r_\gamma r_\delta}\neq\pi_{r_\gamma r_\delta}g_\beta$, then we will set the equation true, by defining
$h(r_\gamma)=r_\delta$
and $h(r_\delta)=r_\gamma$;
if otherwise, we will set the equation false, by
defining $h(r_\gamma)=r_{\delta+1}$
and $h(r_{\delta+1})=r_\gamma$;
in both cases set $h(r_\xi)=r_\xi$
for all other $\xi\in[\alpha_\beta,\delta+1]$ where we haven't yet defined $h(r_\xi)$. Then set $\alpha_{\beta+1}=\delta+2$. Note that this ensures
that $E_h\neq E_{g_\beta}$.
At limit $\beta$ define $\alpha_\beta=\sup_{\beta'<\beta}\alpha_{\beta'}$.
This constructs some $h\in G$,
but $E_h\neq E_{g_\beta}$ for all $\beta<\aleph_1$, which is a contradiction.
