Cayley subgraph isomorphism and complexity of linear subcode decision Let $G$ be an undirected Cayley graph over an abelian group. Let $H$ a regular graph whose independence number and chromatic number are known. Let $inj(G,H)$ be the number of injective homomorphisms from $G$ to $H$. It is known SUBGRAPH ISOMORPHISM is NP-complete. Consider the CAYLEY SUBGRAPH ISOMORPHISM problem: 

Given an undirected Cayley $G$ and a regular $H$, is inj(G,H)>0?

Is the above problem NP-complete?

Since linear codes are abelian, I am extending the question. Given two $[n_i,k_i,d_i]$ linear codes $C_i$ for $i=1$ and $2$ and $n_1&lt;n_2$, is deciding $C_1\cong D \subset C_2$ NP-complete? Note that code isomrphism is similar to graph isomorphism.
For your first question: yes, it is NP-complete. The complete graph is the Cayley graph of any group, if you take the entire group (say, except the identity, but that would just add a self-loop at every vertex) as the generating set. Then CLIQUE is a particular instance of your problem. 

A more interesting question might be to restrict Cayley subgraph isomorphism to Cayley graphs where only a minimal, or say $O(\log|G|)$-sized, generating set is used.

For your second question, the answer is also yes, because not only is Code Equivalence similar to Graph Isomorphism, but GI reduces to Code Equivalence. Essentially the same reduction (due to Petrank and Roth) gives a reduction from Subgraph Isomorphism to Subcode Equivalence.