On the Wikipedia page of the Petersen graph it is mentioned that it is not a Cayley graph.
How it this proved?
Honestly I don't even know how to start this. The only criteria I can think of is that all vertices must have the same degree. Also the degree being odd should imply that one generator of the group has order 2. But then how do I proceed?
Edit: As already mentioned in the comments, the link did not answer my question.
$G = C_{10}$ is abelian, and so if $a$ and $b$ are two of the generators in the Cayley graph then $a^{-1}b^{-1}ab$ gives a cycle of length $4$, but the Petersen graph has none such.
So suppose that $G= D_{10}$. As you say, at least one of the generators, say $a$ must have order $2$. If there is a generator $b$ of order $5$, then $(ab)^2 = 1$, so again we should have a cycle of length $4$, but there is none.
The only other possibility is that all generators have order $2$, but then there is no product of $5$ generators equal to the identity. But the Petrsen graph does have $5$-cycles, contradiction.