(After I finished writing the paragraphs below, I realized that the assumption that the affine step was a partial matching (paths only of length 1) was wrong. I suggest studying the model anyway with a subset of the affine step represented as a matching, and then relaxing this to K steps where J of them are matchings, and see how the estimates change. My guess is that when J is more than K/2, the quality of the estimates does not change much and the answer will still be no, the connected components will not grow significantly.)

Short (non-)answer: I don't know.

Long (also non-) answer: I suspect not, based on heuristics involving random partial matchings in graphs. In addition to other computer experiments to try, I suggest modelling the problem along the lines below.

The affine step for a given relation can be thought of as a partial matching between some vertices in a graph. Instead of walking to infinity, let us consider how large a connected component can be when using a graph with N primes and picking K many partial matchings. I will freely ignore multiplicative constant factors and make reasonable but unverified estimates in this model.

Each step thus is represented by about N/(log N) edges in the graph. (If Hardy and Littlewood can guess at the number of twin prime pairs, I can assume 1/log N fraction of edges with likely the same reasoning and different numeric, but don't take my word for it. A reasonable experiment would see how reality contrasts with this estimate.) So a subset of vertices about the 1/log N the size of the whole graph is involved in the matching. If we wanted to connect the whole graph, we would need K to be at least log N.

If we were choosing subsets at random, two of these subsets would have an intersection about 1/(log N)^2 the size of the whole set, and for K of these the size of the common intersection would vary widely but would be expected to be about 1/(log N)^K, for K reasonably small. The union would be expected to be about a little less than K*(log N) in size.

For K=2, note that if we want a path longer than length 2, one of the edges of the matching has to belong to the intersection of (the subsets of vertices involved in) the two matchings. This is less likely, and I imagine that of the vertices in the intersection, much less than 1/log N match one another. While longer paths are possible, I expect adding an edge decreases the count of such paths of that length by a factor of log N or more.

My guess is that for K fixed, increasing N drops the likelihood of a path of length L to zero.

Gerhard "Start With Assuming No Iterates" Paseman, 2018.05.21.