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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?2012-11-14
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    http://en.wikipedia.org/wiki/Equivariant_map This is the one I am using.2012-11-14
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    I see: what I'd call "a $\,G-$ map". Thanks.2012-11-14
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    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.2012-11-14
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    @DerekHolt if $H=1$ then we're considering $\mathrm{Aut}_G(G)$, which is comprised of right-multiplication-by-$g$ maps for every $g\in G$, so is certainly not trivial (unless $G$ is). Not sure how primitivity is relevant. If $X$ is primitive, then it is $\cong G/H$ for some $H\le G$ with no subgroup between $H$ and $G$; if in addition $H=1$ then $G$ is simple and cyclic hence $\Bbb Z_p$ for a prime $p$, and the regular action of $\Bbb Z_p$ on itself is in fact by $\Bbb Z_p$-set automorphisms (since it's abelian).2016-08-12

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