It is well-known that the Paley graph $P(q)$ is a strongly regular graph with parameters 
$(4t+1,2t,t-1,t)$. Suppose that $v$ is a vertex in the Paley graph $P(q)$. Suppoe that $C$ is the set of all neighbours of $v$ and $D$ be the complement of $C$. 
If one applies Godsil-Mckay switching, then under what conditions
 the new graph is non-isomorphic to the Paley graph $P(q)$?



Godsil-Mckay switching applied on the Paley graph
If you apply what I call local switching to $C$, then the graph that results can be obtained by Seidel switching as follows. Let $Y$ be the Paley graph with an extra isolated vertex.
Apply Seidel switching to set $C$, to get a new graph $Y'$. Then $Y'$ still has an isolated vertex, the remaining vertices induce a graph isomorphic to $P(q)$. (This is a property of the switching class of $P(q)\cup K\_1$.)


The key here is that in $P(q)$, each vertex not in $C\cup\nu$ is joined to exactly half the vertices in $C$.


For the benefit of outsiders, if $C$ is subset of the vertices of a graph $X$ the Seidel switching on $C$ produces the graph $Y$ on the same vertex set, where if $x\in C$ and
$y\notin C$ then $xy$ is an edge in $Y$ if and only if it is not an edge in $X$.
If $X$ and $Y$ are regular with the same valency, they are cospectral.


For local switching choose $C$ so that the subgraph it induces is regular and each vertex not in $C$ is adjacent to all, none, or exactly half the vertices in $C$. Now for each vertex half-joined to $C$, delete those edges joining it to $C$ and the the complementary set. Again the new graph is cospectral to the original graph, and is often not isomorphic to it. (It's going to be really embarrassing if I made a mistake in this description.)


