Given graphs $ X$ and $ Y$ , their *modular product*, which I will denote by $ X \diamond Y$ , has vertex set $ V(X) \times V(Y)$ where two vertices $ (x,y)$ and $ (x’,y’)$ are adjacent if ($ xx’ \in E(X)$ and $ yy’ \in E(Y)$ ) or ($ xx’ \in E(\overline{X})$ and $ yy’ \in E(\overline{Y})$ ), where $ \overline{X}$ and $ \overline{Y}$ are the complements of $ X$ and $ Y$ . Probably the most well known property of this product is that the cliques correspond to common induced subgraphs of $ X$ and $ Y$ , but I am interested in the automorphisms of this product.

If $ \sigma_1$ and $ \sigma_2$ are automorphisms of $ X$ and $ Y$ respectively, then it is not difficult to see that the map $ (x,y) \mapsto (\sigma_1(x), \sigma_2(y))$ is an automorphism of $ X \diamond Y$ . Let’s call this the “Simple Construction”. I am wondering when the only automorphisms of $ X \diamond Y$ are those from the Simple Construction.

A necessary condition for this is that the graphs are not isomorphic. This is because if $ \varphi_1 : V(X) \to V(Y)$ and $ \varphi_2 : V(Y) \to V(X)$ are isomorphisms of $ X$ and $ Y$ , then the map $ (x,y) \mapsto (\varphi_2(y), \varphi_1(x))$ will be an automorphism of $ X \diamond Y$ , and this cannot be obtained by the Simple Construction, (the $ X$ -coordinate of the image depends only on the $ Y$ -coordinate of the input for this construction).

Moreover, if $ \varphi_1$ and $ \varphi_2$ are isomorphisms of $ X$ and $ \overline{Y}$ instead, then the construction above still gives an automorphism of $ X \diamond Y$ . Thus $ X$ and $ \overline{Y}$ being non-isomorphic is also a necessary condition for all automorphisms of $ X \diamond Y$ to be obtainable from the Simple Construction.

It seems that these two conditions are still not sufficient though. I tested pairs of random regular graphs (of random degrees) on 20 vertices in Sage and found a few examples of graphs $ X$ and $ Y$ such that $ X \not\cong Y$ and $ X \not\cong \overline{Y}$ but $ X \diamond Y$ had more automorphisms than those from the Simple Construction. In each of these examples at least one of $ X$ and $ Y$ was a disjoint union of cycles (and the other was of the same form or cubic).

So my question is what additional conditions can I impose on $ X$ and $ Y$ that will ensure that the only automorphisms of $ X \diamond Y$ are those from the Simple Construction? I would prefer something simple such as requiring that $ X, Y, \overline{X}, \overline{Y}$ are all connected. Alternatively, are there any good references for automorphisms of this product?