Problem:
Find a graph $G$ that follows these conditions:
i) The number of vertices is 25
ii) $G$ and it's complement have the same degree sequence
iii) All vertices of $G$ are of the same degree
iv) $G$ is not isomorphic to it's complementArgument why the graph you found follows the conditions, especially condition iv)
My thoughts:
I chose a complete graph on 25 vertices, which has 300 edges. But now I need to somehow divide these 300 edges into a graph $G$ and it's complement $G'$. But I'm not really sure how to progress further, on how to divide these edges.
What steps should I take next?

Take a cycle $C$ on the set of $25$ vertices. Number the vertices $x_0, x_1, x_2,\ldots, x_{24}$ in the order of the cycle; $x_ix_j$ is an edge in $C$ iff $i$ is in $\{j-1,j+1\}$ addition done mod $25$.
Let $G$ be the graph where two vertices $x$ and $y$ are adjacent in $G$, iff the distance from $x$ to $y$ in $C$ is no larger than $6$. Then $G$ is $12$-regular. This forces the complement $G^c$ of $G$ to also be $12$-regular i.e., $G$ and $G^c$ have identical degree sequences.
Note that $G$ does not have an independent set with $4$ vertices. But the complement $G^c$ of $G$ does; $x_1,x_2,x_3,x_4$ form an independent set in $G^c$. So $G$ and $G^c$ cannot be isomorphic to each other.