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Problem So let $\phi : G \rightarrow G'$ be an isomorphism between two directed graphs. Prove that $\phi^{-1}$ is an isomorphism. Also prove if $H \leq Aut(G')$ then $\phi^{-1}H\phi\leq Aut(G)$.

My solution So two digraphs $G,G'$ are isomorphic, means that $\phi:V(G)\rightarrow V(G')$ is a bijection between vertex sets of $G,G'$ and $(u,v)$ is an arc in $G$ if and only if $(\phi(u),\phi(v))$ is an arc in $G'$. Hence to show that $\phi^{-1} : G' \rightarrow G$ is a digraph isomorphism we need to show that $(u,v)$ is an arc in $G'$ if and only if $(\phi^{-1}(u),\phi^{-1}(v))$is an arc in G. Since obviously inverse of a bijection between vertex sets is a bijection as well. So $(u,v)$ is an arc in $G'$ $\Leftrightarrow$ $(\phi\phi^{-1}u,\phi\phi^{-1}v)$ is an arc in $G'$ $\Leftrightarrow$ $(\phi(\phi^{-1}u),\phi(\phi^{-1}v))$ is an arc in $G'$ $\Leftrightarrow$ $(\phi^{-1}u,\phi^{-1}v)$ is an arc in $G$. Hence we showed what we wanted to show.

Now I am stuck in the second part of the problem. Should I simply show all the subgroup properties or the is some other way?

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