Let $G$ be a graph all of whose edge-deleted subgraphs are isomorphic. Is $G$ necessarily edge-transitive?
I checked the following link but in the end they conclude without good justification what happens if the graph is bipartite. Could you please help me? I have a hard time understanding this exercise.
I think the argument is true.
Following the aforementioned claim(hyperlink), we may assume that $G$ is [a bipartite graph with a bipartition $(A,B)$ and $\deg a = d+1$ and $\deg b = d$ for all $a\in A$, $b\in B$] $\cdots(*)$. By the given condition that all edge-deleted subgraphs are isomorphic, we can describe $G-e$ where $e\in E(G)$. Denote $e=vw$ with $v\in A$ and $w\in B$. There are three types of vertices of $G-e$: $\deg = d+1$, $d$, $d-1$. Of course, $\deg_{G-e} v = d$ and $\deg_{G-e} w = d-1$. Remind that $N(v) \subseteq B$, where $N(v)$ is the set of neighbors of $v$.
If we add an edge $f$ in $G-e$ which is incident with $w$, there is only one option that $G-e+f$ is isomorphic to $G$, $f=e$. Otherwise, $G-e+f$ cannot satisfy (*). It implies that $G$ is edge-transitive.