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Couple of days ago, i actually asked this question : Existence of a normal subgroup with $|\operatorname{Aut}{(H)}|>|\operatorname{Aut}{(G)}|$.

I was thinking about the converse of this statement. Suppose $G$ is a finite group, with subgroups $H$ and satisfies $$|Aut(G)|=|Aut(H)|,$$ then can anything be said about, $H$ or $G$ in terms of their structure. Will $H$ be normal or ....? Or one can even ask, that find all finite groups $G$ such that $|Aut(H)|=|Aut(G)|$ for every subgroup $H$ of $G$! I am not interested that much in this question as there seems one can have lot of Groups of this type, but i am more curious to know the behaviour of $H$.

Also, anyone interested can very well read this article in MO: https://mathoverflow.net/questions/1075/order-of-an-automorphism-of-a-finite-group

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    I'd be amazed if there was anything general that could be said about this situation. But here's an example where it holds. For most values of $n$, $S_n$ and $A_n$ have the same order of automorphism group.2010-08-16
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    In the group $C_2\times C_4 \times S_3$, the subgroup $H=C_2\times C_4\times C_2$ satisfies this property, but is not normal.2010-08-16
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    @Robin Chapman: Please provide me with proof of that result, between $S_{n}$ and $A_{n}$.2010-08-17
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    @Chandru1: See pages 299ff in Suzuki's "Group Theory", vol. 1.2010-08-17
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    @Steve D: For my problem or the one which Robin Chapman had stated.2010-08-17
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    @Chandru1: I mean a proof of Robin's statement. I doubt very much there is anything interesting to say about such group/subgroup pairs, as relates to your original question.2010-08-17
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    @Steve D: Ok steve, thanks. If supposing i invoke, some condition do you think that it can be solved. For example $H$ is abelian....something like that2010-08-18

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