It is a known result that if $H$ is the only subgroup with order $|H|$ of a group $G$, then $H$ is characteristic in G (this means that $f(H)=H, \forall f\in{\rm Aut}(G)$).
I was wondering if the following statement is true:
If $H$ is a subgroup of $G$ such that there is no other subgroup isomorphic to $H$, then $H$ is characteristic in G.
How I would prove it:
Let $f\in{\rm Aut}(G)$, then $f|_H:H\longrightarrow f(H)$ is an isomorphism between $H$ and $f(H)$. Since $f(H)$ is a subgroup of $G$ and there is no other subgroup isomorphic to $H$, we have that $f(H)=H$. Therefore, $H$ is characteristic in $G$.
Would this be correct? I am asking because I have not seen it written anywhere, so maybe there is something that I am missing and it is not true.
If this property were true, I would use it here: in $D_6$, there are three subgroups of order $6$, one of them isomorphic to $C_6$ and the other two isomorphic to $S_3$. With the proven result I could say that the group isomorphic to $C_6$ is characteristic in $D_6$.