How do I go about checking if the graph with the given paramaters is a Cayley graph?

A graph is Cayley if and only if its automorphism group has a regular subgroup. A computer algebra system such as Magma or GAP (I think) usually can determine if a permutation group of order 288000 has a regular subgroup rather quickly. If you can give the graph or the automorphism group in some common format, somebody might even run the computation for you. 


There isn't any good general computational method for determining whether a permutation group has a regular subgroup. It was recently described to me by an authority on permutation group algorithms as an important unsolved problem. However, degree 50 is quite small and, as verret noted, various ad hoc methods available in algebra packages are sure to be successful. However, the information you have provided is impossible for a vertextransitive graph regardless of whether or not it is a Cayley graph. Suppose it is vertextransitive. Since 288000 is divisible by 3, there is an automorphism $g$ of order 3. The order 50 is not divisible by 3. Therefore there is a vertex $v$ fixed by $g$ that is adjacent to a vertex moved by $g$. (I didn't assume the graph is connected here.) This means that the 7vertex neighbourhood graph of $v$ has an automorphism of order 3, but the neighbourhood you told us in your comment (a pentagon and two isolated vertices) does not have such an automorphism. Therefore your graph is not a Cayley graph. Unless you got the group order wrong? The largest plausible group order for a connected transitive graph with that neighbourhood graph is 1000. EDIT: Verret's comment to this answer suggested a simpler argument. If there are 5 triangles on each vertex and it has 50 vertices, the total number of triangles must be $\frac{50\times 5}{3}$ which is integralitychallenged. 

