About using conjugation in $S_4$ to help in finding the outer automorphisms of $D_4$

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I'm studying the ${\rm Aut}(D_n)$ from other posts here, even if not all of them are so easy to me but I'm trying to collect them all and get a good view about them.

It's clear ${\rm Inn}(D_4)\cong D_4/Z(D_4)$, which leads to to a group isomorphic to $V_4$. So there will be then remaining four outer automorphisms leading to $Aut(D_4)\cong D_4$.

Considering $D_4$ is isomorphic to a subgroup of $S_4$, I was wondering if using the conjugation of $D_4$ in $S_4$ could give any hints to estimate the other four outer automorphisms.

It's just an intution, but I may be wrong of course.

Thanks in advance.

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I think conjugations in $S_4$ will not help to find all automorphisms of the group $D_4$. But to calculate all automorphisms it is indeed useful to know that the subgroup $$ H=\operatorname{gr}((1324),(12)(34))\cong D_4. $$ For the group $H$ there are the following $4$ pairs of generating involutions: \begin{eqnarray*} (12),(13)(24);\\ (12),(14)(23);\\ (34),(13)(24);\\ (34),(14)(23). \end{eqnarray*} Since every automorphism of the group $H$ is obviously uniquely defined if we know the image of any of these pairs. Here are 8 automorphisms: \begin{eqnarray*} \phi_1(12)=(12),\phi_1((13)(24))=(13)(24);\\ \phi_2(12)=(12),\phi_2((13)(24))=(14)(23);\\ \phi_3(12)=(34),\phi_3((13)(24))=(13)(24);\\ \phi_4(12)=(34),\phi_4((13)(24))=(14)(23);\\ \phi_5(12)=(13)(24),\phi_5((13)(24))=(12);\\ \phi_6(12)=(14)(23),\phi_6((13)(24))=(12);\\ \phi_7(12)=(13)(24),\phi_7((13)(24))=(34);\\ \phi_8(12)=(14)(23),\phi_8((13)(24))=(34);\\ \end{eqnarray*} The first four automorphisms are inner, the last four are outer.