It does seem like there ought to be a simpler solution just based on Hensel's lemma, sort of along OP's reasoning. The given solutions $0, 1,2,3,4,5, 10$ each approximately lift to a 2-adic solution to $f(x) = 0$. Any other $x$ such that $f(x) = 2^n$ for some large $n$ would necessarily lift to one of those same $2$-adic solutions, which means that $x$ would be $2$-adically close to one of the known solutions.

@DanielAsimov Clearly "infinitely many" was intended.

Thanks for the link. It seems a theorem of Whitney describes when $\operatorname{Aut}(G) \to \operatorname{Aut}(L(G))$ fails to be injective or surjective. Answer: It is injective unless $G$ has a $K_2$ component or at least two isolated vertices. It is surjective unless $G$ has both $W_1$ and $W_2$ as components or a component isomorphic to $W_3$, $W_4$, or $W_5$. Here $W_1, \dots, W_5$ are explicit graphs on at most $4$ vertices, namely $W_1 = K_{1,3}$, $W_2 = K_3$, $W_3 = $ a triangle with a pendant edge, $W_4 = K_{1,1,2}$, and $W_5 = K_4$.

Although Q1 itself was too naive, it seems to be a source of somewhat interesting questions to take a given injective functor $F$ from graphs to graphs and ask for which graphs $F:\operatorname{Aut}(G) \to \operatorname{Aut}(F(G))$ fails to be an isomorphism.

The line graph functor also does not preserve the automorphism group. For example, $L(K_4)$ is the complement of $K_2 \sqcup K_2 \sqcup K_2$, which has automorphism group $C_2^3 \rtimes S_3$, of order $48$.

I don't understand. A cycle is non-rigid. Do you mean subdividing a rigid graph doesn't seem to create automorphisms?

Each element of $G$ induces a map $H \to \mathbb R / \mathbb Z$, and this map $G \to H^*$ factors through $G/H$ since the bilinear form is zero on $H$. Assuming the bilinear form is nondegenerate, the image of $G/H \to H^*$ separates the points of $H$, so it is a surjective homomorphism and it follows that $G/H$ is canonically isomorphic to $H^*$ since $|G/H| = |H| = |H^*|$.

Do you assume the bilinear form is nondegenerate? If not, take $q = 0$ and you get easy counterexamples.

The reason the 3-variable DP is not any better is that it follows from the 2-variable DP: $f(x+y+z) \le 2f(x) + 2f(y+z) + 1 \le 2f(x) + 4f(y) + 4f(z) + 3$.