For a simple connected Graph $G$ with $n$ vertices and $m$ edges I want to deduce the number of induced subgraphs $G_k$ on $k$ vertices which are not isomorphic. This boils down to the question of how many copies of one induced subgraph $G_k$ can I find in $G$. Assume that the vertex set of $G_k$ is $V_k$. Then any automorphism $A$ on $G$ satisfying $A(V_k)\neq V_k$ defines a copy of $G_k$ in $G$. Now if $G_k$ and $G_k'$ are isomorphic there exists an automorphism $A$ of $G$ satisfying $A(V_k)\neq V_k$. Hence, any induced subgraph $G_k$ has $a_{G_k}:=|\{A\in Aut(G)|A(V_k)\neq V_k\}|$ copies in $G$. Is this correct so far?
In total we have $\binom{n}{k}$ induced subgraphs on $k$ vertices. How can I combine this with the previous result (if it is correct)? Or is there a formula for this?
EDIT: The claim that I can construct an automorphism from an isomorphism of induced subgraphs is wrong. If I consider for example the circle on 6 vertices $C_6$, labeld from 1 to 6 counterclockwise, and consider the induced subgraphs $G_k = (\{2,5\}, \{\})$ and $G_k = (\{3,5\}, \{\})$ then $I$ which maps $I(3) = 5$ and $I(5) = 2$ defines an isomorphism between the two. Now assume that I can extend $I$ to an automorphism on $G$. Then $I(4)$ is neighbor of both $I(3)=5$ and $I(5)=2$ but $5$ and $2$ do not have a common neighbor in $C_6$.