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:
If each $v \in V(G) $ has the same number of diametral paths initiated from it, then $G$ is a regular 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 $y$.
The answer to your question is $\it{NO}$. The below graph is one example. For each vertex, there is $4$ diametral path, but the graph is not regular.
By my comment in the above, we have another possibility for viewing your problem. I thought that I found a counterexample also for this case, but Tony Huynh noticed me the mistake. I think with this definition, the answer to your question is positive.
Here's another generalization of Tony Huynh's nice construction where every vertex has the same number of diametral paths starting from it and the set of degrees of the vertices in a graph may be large. (The definition of diametral path used here is the one where there is no shorter path between the vertices.) Let's start with a tree of height $h$ where the root has degree $2$ and every vertex in the tree except the root and the leaves has degree $i+3$ where $i$ is the depth of that vertex. The tree has $(h+1)!$ leaves. Take two of such trees and glue their leaves together so that each leaf from the first tree has exactly one edge connecting it to a leaf from the other tree. Such glued trees seem to appear in the contexts of quantum walks. Using the original cycle of length $2k$, replace each edge in the cycle by this glued tree where the roots of the glued tree are replaced by the vertices of the original cycle. It seems that this graph has diameter $(2h+1)k$ and each vertex has $2((h+1)!)^k$ diametral paths starting from it.
