Here is a simple and "natural" (whatever it means) way to get a regular graph from a given graph $G$. Take two disjoint copies of $G$, and insert edges between all pairs of vertices $(v_1,v_2)$ (with $v_1$ from the first copy and $v_2$ from the second copy) whenever there is no edge between the two corresponding vertices of $G$. Denoting by $n$ the order of the original graph $G$, this way you get an $n$-regular graph of order $2n$.

Yet another possible construction. Take four copies of the vertex set of $G$, and insert an edge between the vertex $v_i$ from the $i$th copy and the vertex $v_{i+1}$ from the $(i+1)$-th copy if either $i$ is even and there is an edge between the two corresponding vertices of the original graph $G$, or $i$ is even and there is no edge between the two vertices of $G$. This yields an $n$-regular $4$-partite graph of order $4n$ which keeps the structure of $G$. More generally, for any even integer $k\ge 4$ you can produce this way a $k$-partite, $n$-regular graph of order $kn$, encoding in a very transparent way the structure of the original graph.

Here is a simple and "natural" (whatever it means) way to get a regular graph from a given graph $G$. Take two disjoint copies of $G$, and insert edges between all pairs of vertices $(v_1,v_2)$ (with $v_1$ from the first copy and $v_2$ from the second copy) whenever there is no edge between the two corresponding vertices of $G$. Denoting by $n$ the order of the original graph $G$, this way you get an $n$-regular graph of order $2n$.

Yet another possible construction. Take four copies of the vertex set of $G$, and insert an edge between the vertex $v_i$ from the $i$th copy and the vertex $v_{i+1}$ from the $(i+1)$-th copy if either $i$ is even and there is an edge between the two corresponding vertices of the original graph $G$, or $i$ is odd and there is no edge between the two vertices of $G$. This yields an $n$-regular $4$-partite graph of order $4n$ which keeps the structure of $G$. More generally, for any even integer $k\ge 4$ you can produce this way a $k$-partite, $n$-regular graph of order $kn$, encoding in a very transparent way the structure of the original graph.

Here is a simple and "natural" (whatever it means) way to get a regular graph from a given graph $G$. Take two disjoint copies of $G$, and insert edges between all pairs of vertices $(v_1,v_2)$ (with $v_1$ from the first copy and $v_2$ from the second copy) whenever there is no edge between the two corresponding vertices of $G$. Denoting by $n$ the order of the original graph $G$, this way you get an $n$-regular graph of order $2n$.

Here is a simple and "natural" (whatever it means) way to get a regular graph from a given graph $G$. Take two disjoint copies of $G$, and insert edges between all pairs of vertices $(v_1,v_2)$ (with $v_1$ from the first copy and $v_2$ from the second copy) whenever there is no edge between the two corresponding vertices of $G$. Denoting by $n$ the order of the original graph $G$, this way you get an $n$-regular graph of order $2n$.

Yet another possible construction. Take four copies of the vertex set of $G$, and insert an edge between the vertex $v_i$ from the $i$th copy and the vertex $v_{i+1}$ from the $(i+1)$-th copy if either $i$ is even and there is an edge between the two corresponding vertices of the original graph $G$, or $i$ is even and there is no edge between the two vertices of $G$. This yields an $n$-regular $4$-partite graph of order $4n$ which keeps the structure of $G$. More generally, for any even integer $k\ge 4$ you can produce this way a $k$-partite, $n$-regular graph of order $kn$, encoding in a very transparent way the structure of the original graph.