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If anyne could give me some help with this, it will be deeply appreciated:

Let $X$ be a set equipped with the action of some group $G$. Denote by $Aut_G(X)$ the set of $G-equivariant$ bijections $f: X \to X$ and take for granted that this is a group under composition.

If we let $H$ be a subgroup of $G$ and let $X = G/H$ equipped with the usual G-action (i.e. left multiplication), is it possible to find an isomorphism between $Aut_G(X)$ and $N_G(H)/H$, with $N_G(H)$ being the normalizer of $H$?

Thank you very much in advance for any help.

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What does "equivariant bijection" mean for you? –  DonAntonio Nov 14 '12 at 3:29
    
en.wikipedia.org/wiki/Equivariant_map This is the one I am using. –  user44069 Nov 14 '12 at 3:37
    
I see: what I'd call "a $\,G-$ map". Thanks. –  DonAntonio Nov 14 '12 at 3:46
    
I may be misunderstanding the question, but why should it be true even for $H=1$? If the action of $G$ on $X$ is primitive then ${\rm Aut}_G(X)$ is trivial. –  Derek Holt Nov 14 '12 at 8:48
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