Let $F$ and $G$ be graphs. Let $sub(F, G)$ denotes the number of subgraphs of $G$ that are isomorphic to $F$, let $inj(F, G)$ denote the number of injective homomorphisms from $F$ to $G$ and let $aut(F)$ be the number of automorphisms of $F$.
I am reading a paper that states that $sub(F, G) = inj(F,G)/aut(F)$ as a proposition, but does not provide a proof as the result is "folklore". The paper also does not provide a reference for finding a proof. Although I feel like the result is intuitive (dividing by $aut(F)$ to avoid double counting subgraphs), I would still like to see a proof of this result for a project that I am working on.
Edit: The paper is called "Counting subgraphs via homomorphisms".
Let's use capital letters to denote sets, ie $Sub(F,G)$ is the set of subgraphs of $G$ isomorphic to $F$, and likewise for $Inj(F,G)$ and $Aut(F)$. $Aut(F)$ acts on $Inj(F,G)$ by precomposition. Basically, the point is that $Sub(F,G)$ correspond to the orbits of this action. Given $f\in Inj(F,G)$, taking the image of $f$ gives a subgraph of $G$ isomorphic to $F$. Clearly precomposing with an automorphism maps to the same subgraph, so each orbit maps to one subgraph. Given two injections mapping to the same subgraph, $f$ and $g$, we see $h = g^{-1} \circ f$ is an automorphism of $F$ such that precomposing $g$ by $h$ gives $f$, meaning that two injections mapping to the same subgraph are in the same orbit. That shows there's a $1-1$ correspondence between orbits of $Aut(F)$ acting on $Inj(F,G)$ and $Sub(F,G)$, so $sub(F,G) = inj(F,G)/aut(F)$.