For the purpose of this question, consider a "vertex rearranging function" for a graph G $=($V$,$ E$)$ is a function $\backslash$phi : V $->$ V that rearranges the vertices of G while preserving the graph's structure. This means that if there is an edge between vertices u and v in G$,$ then there must also be an edge between $\backslash$phi $($u$)$ and $\backslash$phi $($v$)$ in G$,$ and if there is no edge between u and v$,$ then there is none between $\backslash$phi $($u$)$ and $\backslash$phi $($v$).$ In other words, $\backslash$phi is a bijective map from V to V such that $($u$,$v$)$ in E implies $(\backslash$phi $($u$),\backslash$phi $($v$))$ in E and $($u$,$v$)$ E implies $(\backslash$phi $($u$),\backslash$phi $($v$))$ E$.$

Show that the graph G $=(\{1,2,3,4\},\{\{1,2\},\{2,3\},\{1,3\},\{3,4\}\})($shown in the stimulus on the left$)$ only has one possible non$-$trivial $($that doesn't simply map each vertex to itself$)$ 'vertex rearranging function' as defined above.