My textbook, Szekeres's "A Course in Modern Mathematical Physics: Groups, Hilbert Space and Differential Geometry", tacitly alludes to the following lemma, which I've tried to write out and prove on my own:
Two subgroups $H$ and $H'$ are said to be conjugate subgroups if there exists\ an inner automorphism such that $H' = C_g(H)$ (that is, if the image of $H$ under that inner automorphism equals $H'$). Then the relation of being conjugate subgroups is an equivalence relation on the set of all subgroups of $G$.
My proof of the symmetry requirement is below, but it feels very long as I've "unravelled" both sides of the set equality. I have to imagine there's some more direct way to use that an inner automorphism is, in particular, a bijection of $G$. Can anyone provide that?
Symmetry: If $H$ is conjugate to $H'$ then, by definition, $H' = C_g(H)$ for some $g \in G$. In particular, $H' \supset C_g(H)$, so for every $h \in H$ there exists an $h' \in H'$ such that $h' = ghg^{-1}$. That is, for every $h \in H$ there exists an $h' \in H'$ such that $g^{-1}h'g = h$. Thus $H \subset C_{g^{-1}}(H')$. Now we also have $H' \subset C_g(H)$, so that for every $h'$ there exists some $h$ such that $h' = ghg^{-1}$. Thus for every for every $h'$ there exists some $h$ such that $g^{-1}h'g = h$ so that $H \supset C_{g^{-1}}(H')$. Thus $H = C_{g^{-1}}(H')$ and so there does indeed exist an inner automorphism making $H'$ conjugate to $H$.