L(G) is isomorphic to G iff G is a cycle The converse is pretty obvious. If G is a cycle,  then it is isomorphic to it's line graph.
How to prove that if L(G) is isomorphic to G, then G is a cycle...?
P.S.- Assume G is connected
 A: A vertex of $G$ of degree $d_i$ contributes $\binom{d_i}2$ edges to $L(G)$. Then with $n$ denoting the common number of vertices and edges of $G$ and $L(G)$,
$$
\sum_id_i=2n\;,
$$
$$
\sum_i\frac{d_i(d_i-1)}2=n\;,
$$
so
$$
\sum_id_i^2=4n
$$
and thus
\begin{align}
\operatorname{Var}(d)&=E[d^2]-E[d]^2
\\
&=\frac{\sum_id_i^2}n-\left(\frac{\sum_id_i}n\right)^2
\\
&=4-4
\\
&=0\;.
\end{align}
Thus all vertices in both graphs have the same degree $2$, which is only the case in a union of cycle graphs.
A: Let $G$ be a connected graph on $n$ vertices and $m$ edges and suppose $G$ is isomorphic to its line graph $L(G)$.  Then, $m=n$ and so $G$ is a connected graph having $n$ edges.  This implies $G$ is of the form $T+e$, where $T$ is a tree and $e$ is an edge not in $T$.  If $G=T+e$ is the $n$-cycle graph, then we are done.  
So suppose $T+e$ is not a cycle.  Then, $T+e$ has a vertex of degree 3 or more. Three of the edges incident to this vertex in $G$ form a cycle of length 3 in $L(G)$.  Also, the unique cycle in $G$ obtained by adding edge $e$ to the tree $T$ induces a cycle in $L(G)$. Hence, $L(G)$ has at least two cycles.  We showed that $G$ has only one cycle and that $L(G)$ has two or more cycles, which is a contradiction because $G \cong L(G)$.  So this case is impossible, and $T+e$ must be a cycle.
