The sum of degrees is $10$, and since the graph is self complementary, by symmetry the degree sequence must be
$$(d_1,d_2, 2,4-d_2,4-d_1) \,,$$
where $d_1, d_2 \in \{ 0,1,2 \}$, and $d_1 \leq d_2$.
It is easy to see that $d_1=0$ is not possible, since then the last degree would be $4$, thus $d_1, d_2 \in \{ 1,2 \}$, and $d_1 \leq d_2$.
Case $1$ $d_1=1, d_2=1$. Your degree sequence is $(1,1,2,3,3)$. Each of the vertices of degree $3$ must be connected to all vertices excluding one end vertex. There is only one graph up to isomorphism.
Case $2$ $d_1=1, d_2=2$. Your degree sequence is $(1,2,2,2,3)$. You have two graphs here: the end vertex is connected to a degree $2$ or $3$.
Case $3$ $d_1=2, d_2=2$. Your graph is connected (why?) and has all degrees $2$. Only 1 possibility.
A sample isomorphism for your graphs is relatively simple here if you know that the Petersen graph is strongly regular, meaning that the choice of starting point for your isomorphism is arbitrary.
So we can map between the two diagrams starting with an arbitrary three-point arc, so taking mapping pairs of $(a,1),(b,2),(c,3)$ we can continue the 5-cycle with $(e,9),(f,10)$ then completing adjacent sets we have $(d,4), (i,5), (k,8), (h,6),(g,7)$ giving a full isomorphism:
$$(a,1),(b,2),(c,3),(d,4),(e,9),(f,10),(g,7), (h,6), (i,5), (k,8)$$
Due to the strong regularity, we can pick any starting point, any of its adjacent points then any further adjacent point to map to a given arc of three points on one graph, giving $10\times 3\times 2= 60$ different isomorphisms to choose from.
Best Answer
The graph on the left is Fanny. You will find the isomorphisms by matching the vertices of degree 2.
The graph on the right isn't any of the four.
Note that OEIS says there are 10 self-complementary graphs on 8 vertices https://oeis.org/A000171. They are shown in http://mathworld.wolfram.com/Self-ComplementaryGraph.html In fact your graph from the right picture appears there as "8-self-complementary graph 3".