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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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    @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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Suppose $f:G/H\to G/H$ is equivariant. As such, it is determined by where it sends $H$, since on the rest of $G/H$ we have $f(gH)=gf(H)$. Say $f(H)=\sigma H$. Then $f(hH)=hf(H)$ implies

$\sigma H=h\sigma H~~\Leftrightarrow~~ H=\sigma^{-1}h\sigma H ~~\Leftrightarrow~~ \sigma^{-1}h\sigma\in H.$

for any $h\in H$. In other words, $\sigma^{-1}\in N_G(H)$, and so $\sigma\in N_G(H)$. Conversely, if $\sigma\in N_G(H)$, then $\sigma H=H\sigma$, so any putative mapping $f(gH)=g\sigma H$ is just $gH\mapsto gH\sigma$, i.e. right multiplication by $\sigma$ which clearly commutes with the left $G$-action and has an inverse.

I leave it as an exercise to show the surjection $N_G(H)\to\mathrm{Aut}_G(G/H)$ is a homomorphism with kernel just $H$. By the first isomorphism theorem, ${\rm Aut}_G(G/H)\cong N_G(H)/N$.