This is more of a comment than an answer, because it seems like I missed something in your question.
First of all, is it possible for such a schedule to be acyclic?
I also wonder about your symmetry constraint. Doesn't that make all such graphs symmetrical to each other? Is it permissible to rearrange rows and columns of a join matrix?
For example, if we allow self-connections in a graph, does the following connection matrix match your conditions?
1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1
Starting with this matrix, is it possible to rearrange rows and columns from it to obtain all such graphs, where all rows and columns have the sum of three?
One example of such a matrix can be obtained from the above matrix A as follows (using MATLAB).
>> A(randperm(8),randperm(8)) ans = 0 1 0 0 0 1 1 0 0 0 1 0 1 0 1 0 1 1 0 1 0 0 0 0 1 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 0 1 0 0 1 0 0 0 0 1 1 1 0
PS. In this case, I repeated the command several times to get a matrix with zeros in the diagonal. :)
Edit
And, I see from your comments that I'm wrong. Of course, the permutation index should be the same for rows and columns. At least I should have noticed this when I started with a schedule with self-connections and got one without them after the permutation.
A random isomorphic permutation will look like this:
idx = randperm(8); A(idx,idx);
which saves all self-connections.
Perhaps this may be useful in creating matrices, but it is not at all as useful as I thought it would be.