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I will prove that if $G \cong \mathrm{Aut}(G)$ and $G$ is finite, then $G$ is the Cayley graph of a finite-dimensional vector space over $\mathbb{Z}/2\mathbb{Z}$. Then with Godsil's theorem about the existence of graphical regular representation (see page 32 of this) there are examples with arbitrary large finite cardinals.

Let's denote $Q(G)$ the automorphism group of any graph $G$. Then $\mathrm{Aut}(G)$ is a graph with vertices indexed by $Q(G)$. Suppose $G \cong \mathrm{Aut}(G)$, one has $\mathrm{Aut}(G) \cong \mathrm{Aut}(\mathrm{Aut}(G))$. However, in $\mathrm{Aut}(\mathrm{Aut}(G))$, $2$ types of vertices can be found:

1. The left translations $\{l_g: \varphi \mapsto g\varphi \mid g \in Q(G)\}$. Given $g \in Q(G)$, $l_g$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{g\varphi(v), g\psi(v)\}\in E \implies \{g\varphi, g\psi\}=l_g\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{\varphi g(v), \psi g(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{\varphi g, \psi g\}= \{l_\varphi(g), l_\psi(g)\} \in E \iff \{l_\varphi, l_\psi\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $L(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{l_g\}$ is isomorphic to $\mathrm{Aut}(G)$.

2. The right translations $\{r_g^{-1}: \varphi \mapsto \varphi g \mid g \in Q(G)\}$. Given $g \in Q(G)$, $r_g^{-1}$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{\varphi g(v), \psi g(v)\}\in E \implies \{\varphi g, \psi g\}=r_g^{-1}\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{g\varphi(v), g\psi(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{g\varphi, g\psi\}= \{r_\varphi^{-1}(g), r_\psi^{-1}(g)\} \in E \iff \{r_\varphi^{-1}, r_\psi^{-1}\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $R(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{r_g^{-1}\}$ is also isomorphic to $\mathrm{Aut}(G)$.

When $G$ is a finite graph, the statement above immediately implies that $L(G) = R(G) = \mathrm{Aut}(\mathrm{Aut}(G)) \cong \mathrm{Aut}(G)$. Considering the image of $e \in Q(G) = V(\mathrm{Aut}(G))$ under the action of $Q(\mathrm{Aut}(G))$, one must have $l_g = r_g^{-1}$ in $Q(\mathrm{Aut}(G)) \cong Q(G)$, which means that $Q(G)$ is commutative.

Denote the set of vertices adjacent to $e$ by $S \subseteq Q(G)$, then since $\{l_g\}$ respects adjacency, the connected component of $e$ in $\mathrm{Aut}(G)$ is the Cayley graph of the group generated by $S$. Connectedness of $\mathrm{Aut}(G)$ implies that $\mathrm{Aut}(G)$ is the Cayley graph of the commutative group $Q(G)$ with generators $S$. $S$ is clearly symmetric: $\{e,g\} \in E(\mathrm{Aut}(G)) \implies l_{g^{-1}}\{e,g\} = \{g^{-1},e\} \in E(\mathrm{Aut}(G))$.

If $Q(G)$ is not a vector space over $\mathbb{Z}/2\mathbb{Z}$, then an element in $S$ must have order $k \ne 2$. Thus $g \mapsto g^{-1}$ is a non-trivial automorphism of $\mathrm{Aut}(G) = \mathrm{Cay}(Q(G),S)$ that is not of the form $\{l_g\}$, contradicting $L(G) = \mathrm{Aut}(G)$. The claim then follows.

I will prove that if $G \cong \mathrm{Aut}(G)$ and $G$ is finite, then $G$ is the Cayley graph of a commutative group of exponent $2$. Then with Godsil's theorem about the existence of graphical regular representation (see page 32 of this, sadly I have no access to the original paper) there are examples with arbitrary large finite cardinals.

Let's denote $Q(G)$ the automorphism group of any graph $G$. Then $\mathrm{Aut}(G)$ is a graph with vertices indexed by $Q(G)$. Suppose $G \cong \mathrm{Aut}(G)$, one has $\mathrm{Aut}(G) \cong \mathrm{Aut}(\mathrm{Aut}(G))$. However, in $\mathrm{Aut}(\mathrm{Aut}(G))$, $2$ types of vertices can be found:

1. The left translations $\{l_g: \varphi \mapsto g\varphi \mid g \in Q(G)\}$. Given $g \in Q(G)$, $l_g$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{g\varphi(v), g\psi(v)\}\in E \implies \{g\varphi, g\psi\}=l_g\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{\varphi g(v), \psi g(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{\varphi g, \psi g\}= \{l_\varphi(g), l_\psi(g)\} \in E \iff \{l_\varphi, l_\psi\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $L(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{l_g\}$ is isomorphic to $\mathrm{Aut}(G)$.

2. The right translations $\{r_g^{-1}: \varphi \mapsto \varphi g \mid g \in Q(G)\}$. Given $g \in Q(G)$, $r_g^{-1}$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{\varphi g(v), \psi g(v)\}\in E \implies \{\varphi g, \psi g\}=r_g^{-1}\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{g\varphi(v), g\psi(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{g\varphi, g\psi\}= \{r_\varphi^{-1}(g), r_\psi^{-1}(g)\} \in E \iff \{r_\varphi^{-1}, r_\psi^{-1}\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $R(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{r_g^{-1}\}$ is also isomorphic to $\mathrm{Aut}(G)$.

When $G$ is a finite graph, the statement above immediately implies that $L(G) = R(G) = \mathrm{Aut}(\mathrm{Aut}(G)) \cong \mathrm{Aut}(G)$. Considering the image of $e \in Q(G) = V(\mathrm{Aut}(G))$ under the action of $Q(\mathrm{Aut}(G))$, one must have $l_g = r_g^{-1}$ in $Q(\mathrm{Aut}(G)) \cong Q(G)$, which means that $Q(G)$ is commutative.

Denote the set of vertices adjacent to $e$ by $S \subseteq Q(G)$, then since $\{l_g\}$ respects adjacency, the connected component of $e$ in $\mathrm{Aut}(G)$ is the Cayley graph of the group generated by $S$. Connectedness of $\mathrm{Aut}(G)$ implies that $\mathrm{Aut}(G)$ is the Cayley graph of the commutative group $Q(G)$ with generators $S$. $S$ is clearly symmetric: $\{e,g\} \in E(\mathrm{Aut}(G)) \implies l_{g^{-1}}\{e,g\} = \{g^{-1},e\} \in E(\mathrm{Aut}(G))$.

If $Q(G)$ is not of exponent $2$, then an element in $S$ must have order $k \ne 2$. Thus $g \mapsto g^{-1}$ is a non-trivial automorphism of $\mathrm{Aut}(G) = \mathrm{Cay}(Q(G),S)$ that is not of the form $\{l_g\}$, contradicting $L(G) = \mathrm{Aut}(G)$. The claim then follows.

Partial Answer: If $G \cong \mathrm{Aut}(G)$ and $G$ is finite, then $G$ is the Cayley graph of a finite-dimensional vector space over $\mathbb{Z}/2\mathbb{Z}$.

Let's denote $Q(G)$ the automorphism group of any graph $G$. Then $\mathrm{Aut}(G)$ is a graph with vertices indexed by $Q(G)$. Suppose $G \cong \mathrm{Aut}(G)$, one has $\mathrm{Aut}(G) \cong \mathrm{Aut}(\mathrm{Aut}(G))$. However, in $\mathrm{Aut}(\mathrm{Aut}(G))$, $2$ types of vertices can be found:

1. The left translations $\{l_g: \varphi \mapsto g\varphi \mid g \in Q(G)\}$. Given $g \in Q(G)$, $l_g$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{g\varphi(v), g\psi(v)\}\in E \implies \{g\varphi, g\psi\}=l_g\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{\varphi g(v), \psi g(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{\varphi g, \psi g\}= \{l_\varphi(g), l_\psi(g)\} \in E \iff \{l_\varphi, l_\psi\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $L(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{l_g\}$ is isomorphic to $\mathrm{Aut}(G)$.

2. The right translations $\{r_g^{-1}: \varphi \mapsto \varphi g \mid g \in Q(G)\}$. Given $g \in Q(G)$, $r_g^{-1}$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{\varphi g(v), \psi g(v)\}\in E \implies \{\varphi g, \psi g\}=r_g^{-1}\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{g\varphi(v), g\psi(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{g\varphi, g\psi\}= \{r_\varphi^{-1}(g), r_\psi^{-1}(g)\} \in E \iff \{r_\varphi^{-1}, r_\psi^{-1}\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $R(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{r_g^{-1}\}$ is also isomorphic to $\mathrm{Aut}(G)$.

When $G$ is a finite graph, the statement above immediately implies that $L(G) = R(G) = \mathrm{Aut}(\mathrm{Aut}(G)) \cong \mathrm{Aut}(G)$. Considering the image of $e \in Q(G) = V(\mathrm{Aut}(G))$ under the action of $Q(\mathrm{Aut}(G))$, one must have $l_g = r_g^{-1}$ in $Q(\mathrm{Aut}(G)) \cong Q(G)$, which means that $Q(G)$ is commutative.

Denote the set of vertices adjacent to $e$ by $S \subseteq Q(G)$, then since $\{l_g\}$ respects adjacency, the connected component of $e$ in $\mathrm{Aut}(G)$ is the Cayley graph of the group generated by $S$. Connectedness of $\mathrm{Aut}(G)$ implies that $\mathrm{Aut}(G)$ is the Cayley graph of the commutative group $Q(G)$ with generators $S$. $S$ is clearly symmetric: $\{e,g\} \in E(\mathrm{Aut}(G)) \implies l_{g^{-1}}\{e,g\} = \{g^{-1},e\} \in E(\mathrm{Aut}(G))$.

If $Q(G)$ is not a vector space over $\mathbb{Z}/2\mathbb{Z}$, then an element in $S$ must have order $k \ne 2$. Thus $g \mapsto g^{-1}$ is a non-trivial automorphism of $\mathrm{Aut}(G) = \mathrm{Cay}(Q(G),S)$ that is not of the form $\{l_g\}$, contradicting $L(G) = \mathrm{Aut}(G)$. The claim then follows.

I will prove that if $G \cong \mathrm{Aut}(G)$ and $G$ is finite, then $G$ is the Cayley graph of a finite-dimensional vector space over $\mathbb{Z}/2\mathbb{Z}$. Then with Godsil's theorem about the existence of graphical regular representation (see page 32 of this) there are examples with arbitrary large finite cardinals.

Let's denote $Q(G)$ the automorphism group of any graph $G$. Then $\mathrm{Aut}(G)$ is a graph with vertices indexed by $Q(G)$. Suppose $G \cong \mathrm{Aut}(G)$, one has $\mathrm{Aut}(G) \cong \mathrm{Aut}(\mathrm{Aut}(G))$. However, in $\mathrm{Aut}(\mathrm{Aut}(G))$, $2$ types of vertices can be found:

1. The left translations $\{l_g: \varphi \mapsto g\varphi \mid g \in Q(G)\}$. Given $g \in Q(G)$, $l_g$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{g\varphi(v), g\psi(v)\}\in E \implies \{g\varphi, g\psi\}=l_g\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{\varphi g(v), \psi g(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{\varphi g, \psi g\}= \{l_\varphi(g), l_\psi(g)\} \in E \iff \{l_\varphi, l_\psi\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $L(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{l_g\}$ is isomorphic to $\mathrm{Aut}(G)$.

2. The right translations $\{r_g^{-1}: \varphi \mapsto \varphi g \mid g \in Q(G)\}$. Given $g \in Q(G)$, $r_g^{-1}$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{\varphi g(v), \psi g(v)\}\in E \implies \{\varphi g, \psi g\}=r_g^{-1}\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{g\varphi(v), g\psi(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{g\varphi, g\psi\}= \{r_\varphi^{-1}(g), r_\psi^{-1}(g)\} \in E \iff \{r_\varphi^{-1}, r_\psi^{-1}\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $R(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{r_g^{-1}\}$ is also isomorphic to $\mathrm{Aut}(G)$.

When $G$ is a finite graph, the statement above immediately implies that $L(G) = R(G) = \mathrm{Aut}(\mathrm{Aut}(G)) \cong \mathrm{Aut}(G)$. Considering the image of $e \in Q(G) = V(\mathrm{Aut}(G))$ under the action of $Q(\mathrm{Aut}(G))$, one must have $l_g = r_g^{-1}$ in $Q(\mathrm{Aut}(G)) \cong Q(G)$, which means that $Q(G)$ is commutative.

Denote the set of vertices adjacent to $e$ by $S \subseteq Q(G)$, then since $\{l_g\}$ respects adjacency, the connected component of $e$ in $\mathrm{Aut}(G)$ is the Cayley graph of the group generated by $S$. Connectedness of $\mathrm{Aut}(G)$ implies that $\mathrm{Aut}(G)$ is the Cayley graph of the commutative group $Q(G)$ with generators $S$. $S$ is clearly symmetric: $\{e,g\} \in E(\mathrm{Aut}(G)) \implies l_{g^{-1}}\{e,g\} = \{g^{-1},e\} \in E(\mathrm{Aut}(G))$.

If $Q(G)$ is not a vector space over $\mathbb{Z}/2\mathbb{Z}$, then an element in $S$ must have order $k \ne 2$. Thus $g \mapsto g^{-1}$ is a non-trivial automorphism of $\mathrm{Aut}(G) = \mathrm{Cay}(Q(G),S)$ that is not of the form $\{l_g\}$, contradicting $L(G) = \mathrm{Aut}(G)$. The claim then follows.

Partial Answer: If $G \cong \mathrm{Aut}(G)$ and $G$ is finite, then $G$ is the Cayley graph of some commutative $2-$group.

Let's denote $Q(G)$ the automorphism group of any graph $G$. Then $\mathrm{Aut}(G)$ is a graph with vertices indexed by $Q(G)$. Suppose $G \cong \mathrm{Aut}(G)$, one has $\mathrm{Aut}(G) \cong \mathrm{Aut}(\mathrm{Aut}(G))$. However, in $\mathrm{Aut}(\mathrm{Aut}(G))$, $2$ types of vertices can be found:

1. The left translations $\{l_g: \varphi \mapsto g\varphi \mid g \in Q(G)\}$. Given $g \in Q(G)$, $l_g$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{g\varphi(v), g\psi(v)\}\in E \implies \{g\varphi, g\psi\}=l_g\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{\varphi g(v), \psi g(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{\varphi g, \psi g\}= \{l_\varphi(g), l_\psi(g)\} \in E \iff \{l_\varphi, l_\psi\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $L(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{l_g\}$ is isomorphic to $\mathrm{Aut}(G)$.

2. The right translations $\{r_g^{-1}: \varphi \mapsto \varphi g \mid g \in Q(G)\}$. Given $g \in Q(G)$, $r_g^{-1}$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{\varphi g(v), \psi g(v)\}\in E \implies \{\varphi g, \psi g\}=r_g^{-1}\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{g\varphi(v), g\psi(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{g\varphi, g\psi\}= \{r_\varphi^{-1}(g), r_\psi^{-1}(g)\} \in E \iff \{r_\varphi^{-1}, r_\psi^{-1}\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $R(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{r_g^{-1}\}$ is also isomorphic to $\mathrm{Aut}(G)$.

When $G$ is a finite graph, the statement above immediately implies that $L(G) = R(G) = \mathrm{Aut}(\mathrm{Aut}(G)) \cong \mathrm{Aut}(G)$. Considering the image of $e \in Q(G) = V(\mathrm{Aut}(G))$ under the action of $Q(\mathrm{Aut}(G))$, one must have $l_g = r_g^{-1}$ in $Q(\mathrm{Aut}(G)) \cong Q(G)$, which means that $Q(G)$ is commutative.

Denote the set of vertices adjacent to $e$ by $S \subseteq Q(G)$, then since $\{l_g\}$ respects adjacency, the connected component of $e$ in $\mathrm{Aut}(G)$ is the Cayley graph of the group generated by $S$. Connectedness of $\mathrm{Aut}(G)$ implies that $\mathrm{Aut}(G)$ is the Cayley graph of the commutative group $Q(G)$ with generators $S$. $S$ is clearly symmetric: $\{e,g\} \in E(\mathrm{Aut}(G)) \implies l_{g^{-1}}\{e,g\} = \{g^{-1},e\} \in E(\mathrm{Aut}(G))$.

If $Q(G)$ is not a $2-$group, then an element in $S$ must have order $k \ne 2$. Thus $g \mapsto g^{-1}$ is a non-trivial automorphism of $\mathrm{Aut}(G) = \mathrm{Cay}(Q(G),S)$ that is not of the form $\{l_g\}$, contradicting $L(G) = \mathrm{Aut}(G)$. The claim then follows.

Partial Answer: If $G \cong \mathrm{Aut}(G)$ and $G$ is finite, then $G$ is the Cayley graph of a finite-dimensional vector space over $\mathbb{Z}/2\mathbb{Z}$.

Let's denote $Q(G)$ the automorphism group of any graph $G$. Then $\mathrm{Aut}(G)$ is a graph with vertices indexed by $Q(G)$. Suppose $G \cong \mathrm{Aut}(G)$, one has $\mathrm{Aut}(G) \cong \mathrm{Aut}(\mathrm{Aut}(G))$. However, in $\mathrm{Aut}(\mathrm{Aut}(G))$, $2$ types of vertices can be found:

1. The left translations $\{l_g: \varphi \mapsto g\varphi \mid g \in Q(G)\}$. Given $g \in Q(G)$, $l_g$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{g\varphi(v), g\psi(v)\}\in E \implies \{g\varphi, g\psi\}=l_g\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{\varphi g(v), \psi g(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{\varphi g, \psi g\}= \{l_\varphi(g), l_\psi(g)\} \in E \iff \{l_\varphi, l_\psi\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $L(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{l_g\}$ is isomorphic to $\mathrm{Aut}(G)$.

2. The right translations $\{r_g^{-1}: \varphi \mapsto \varphi g \mid g \in Q(G)\}$. Given $g \in Q(G)$, $r_g^{-1}$ is clearly a bijection on $V(\mathrm{Aut}(G))$. To see it also respects adjacency, observe that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \implies \forall v \in V, \{\varphi(v), \psi(v)\}\in E \overset{g \in Q(G)}{\implies} \forall v \in V, \{\varphi g(v), \psi g(v)\}\in E \implies \{\varphi g, \psi g\}=r_g^{-1}\{\varphi, \psi\} \in E(\mathrm{Aut}(G))$$ One can also get that $$\{\varphi, \psi\} \in E(\mathrm{Aut}(G)) \iff \forall v \in V, \{\varphi(v), \psi(v)\} \in E \iff \forall v \in V, g \in Q(G), \{g\varphi(v), g\psi(v)\} \in E \iff \forall g \in Q(G)=V(\mathrm{Aut}(G)), \{g\varphi, g\psi\}= \{r_\varphi^{-1}(g), r_\psi^{-1}(g)\} \in E \iff \{r_\varphi^{-1}, r_\psi^{-1}\} \in E(\mathrm{Aut}(\mathrm{Aut}(G)))$$ Therefore, the induced subgraph $R(G)$ of $\mathrm{Aut}(\mathrm{Aut}(G))$ with vertices $\{r_g^{-1}\}$ is also isomorphic to $\mathrm{Aut}(G)$.

When $G$ is a finite graph, the statement above immediately implies that $L(G) = R(G) = \mathrm{Aut}(\mathrm{Aut}(G)) \cong \mathrm{Aut}(G)$. Considering the image of $e \in Q(G) = V(\mathrm{Aut}(G))$ under the action of $Q(\mathrm{Aut}(G))$, one must have $l_g = r_g^{-1}$ in $Q(\mathrm{Aut}(G)) \cong Q(G)$, which means that $Q(G)$ is commutative.

Denote the set of vertices adjacent to $e$ by $S \subseteq Q(G)$, then since $\{l_g\}$ respects adjacency, the connected component of $e$ in $\mathrm{Aut}(G)$ is the Cayley graph of the group generated by $S$. Connectedness of $\mathrm{Aut}(G)$ implies that $\mathrm{Aut}(G)$ is the Cayley graph of the commutative group $Q(G)$ with generators $S$. $S$ is clearly symmetric: $\{e,g\} \in E(\mathrm{Aut}(G)) \implies l_{g^{-1}}\{e,g\} = \{g^{-1},e\} \in E(\mathrm{Aut}(G))$.

If $Q(G)$ is not a vector space over $\mathbb{Z}/2\mathbb{Z}$, then an element in $S$ must have order $k \ne 2$. Thus $g \mapsto g^{-1}$ is a non-trivial automorphism of $\mathrm{Aut}(G) = \mathrm{Cay}(Q(G),S)$ that is not of the form $\{l_g\}$, contradicting $L(G) = \mathrm{Aut}(G)$. The claim then follows.