$G\cong L(G)$ iff there is a $r\in \mathbb{N}$ such that $G=C_{i_1}+C_{i_2}+...+C_{i_r}$ , $i_j\ge3$ and $1\le j\leq r$; equivalently, iff for some positive integer $r$, the graph $G$ is a collection of $r$ vertex-disjoint cycles $C_{i_1}, C_{i_2}, \ldots, C_{i_r}$.
I think I have to use that $C_r, r\ge3$ is transitive this means that there is an automorpism $Aut(C_r): s(x)=y$, $x,y \in V(C_r)$
But I'm not really sure how, any ideas on that?
I am assuming you mean $G$ is the disjoint union of a set of cycles? Because of this I am just going to assume $G$ is connected to speed things up, the proof for $G$ being disconnected will follow immediately.
We note fairly quick that if $G=C_n$ then $L(G) \cong C_n$ from the definition of a line graph.
Suppose $G \cong L(G)$. By the definition of a line graph, $|V(L(G))|=|E(G)|.$ As $L(G) \cong G$, we also have $|V(L(G))|=|V(G)|,$ thus $|V(G)|=|E(G)|$. As $G$ is connected this implies that $G$ contains exactly one cycle.
We now need to prove that every vertex of $G$ has degree 2. We first note that $G$ cannot have any degree 4 or higher vertices as they will generate cliques of size 4 or higher (and hence extra cycles). Next, we note that the number of vertices of degree 1 must equal the number of vertices of degree 3; this follows as every degree 3 vertex gives a branching path from our main cycle, and every one of these paths must end in a vertex of degree 1 (this can be proved more rigorously by a basic induction argument). Given that $V_n$ denotes the degree $n$ vertices of $G$, and recalling that $|V_1 | = |V_3|$, the amount of edges in the line graph of $G$ is given by the equation $$ |E(L(G))| = \sum_{n=2}^\infty \binom{n}{2} |V_n| = |V_2 |+ 3 |V_3| \geq |V_1| + |V_2 |+ |V_3| = |V(G)|,$$ with equality if and only if $|V_1| = |V_3| = 0$. Hence every vertex of $G$ has degree 2 as required.