Let $G(V,E)$ be a graph. A path whose length is equal to the diameter of a graph is called a **diametral** path. In a cycle graph every vertex has $2$ diametral paths. Now I need to prove that this:



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> If each $v \in V(G) $ has the same number of diametral paths initiated from it, then $G$ is a regular graph.
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Relation between diametral path and regularity of a graph
The claim is actually **false**, for both versions of the problem. That is, we may define a *diametral path* of $G$ as a path with length equal to the diameter of $G$ (I think this is the OP's intent), or as a path with length equal to the diameter with the additional property that it is a shortest path between its ends (this is Shahrooz Janbaz's interpretation). 


Here is an infinite family of counterexamples to the second definition (Shahrooz Janbaz has already given a counterexample to the first defintion). Let $G\_k$ be the graph obtained from a cycle of length $2k$ by first adding a parallel edge for each edge and then subdividing each edge once. Observe that $G\_k$ is not regular (it has vertices of degree $2$ and $4$) and has diameter $2k$. Moreover, if $x$ is a degree $4$ vertex of $G\_k$, then there are $2^{k+1}$ diametral paths starting at $x$ (there are $2^k$ in each direction). Similarly, if $y$ is a degree $2$ vertex of $G\_k$, there are $2^{k+1}$ diametral paths starting at $y$. 


