$Aut_G(G/H)$ is isomorph to $N_G(H)/H$ If X is a G-set, an automorphism $G$-invariant is a map $f:X \to X$ such that $f(g*x)=g*f(x)$
I need to prove that $N_G(H)/H$ is isomorphic to $G$-automorphism invariants of $G/H$
I tried with two natural maps:


*

*The map such that $\phi(gH)=f_{gH}$ where $f_{gH}(g'H)=(g'g)H$, but this map isn't homomorphish, and 

*$\phi(gH)=f_{gH}$ where $f_{gH}(g'H)= (gg'g^-1)H$, but this map isn't $G$-invariant.
please, can you give me some ideas?
 A: We have to define a homomorphism $\varphi:N_G(H)\to \mathrm{Aut_G}(G/H)=\{G$-invariant automorphisms of $G/H\}$ and show $\ker\varphi=H$.
Your first applicant works (provided you write composition [of automorphisms] to the right): $\phi(x):= gH\mapsto gxH$.  
Else, we have to take $\varphi(x):=\phi(x^{-1})=gH\mapsto gx^{-1}H$.
Note that $x\in N_G(H)$ means $xH=Hx$, so $gxH=gHx$ and these maps are well defined: if $gH=g'H$, then $gxH=gHx=g'Hx=g'xH$, and are also $G$-invariant: 
$$\phi(x)(g\cdot uH)= guHx =g\cdot\phi(x)(uH) $$
Now it's not $\phi(x)$ that needs to be a homomorphism, but it's $\phi$ [or $\varphi$, depending which direction the functions are composed].
$$\varphi(x)\big(\varphi(y)(uH)\big)=uHy^{-1}x^{-1}=uH(xy)^{-1}=\varphi(xy)(uH)$$
$$(uH^{\phi(x)})^{\phi(y)}=(uHx)y=uHxy=(uH)^{\phi(xy)}$$
Finally, $\phi(x)=\mathrm{id}$ implies $\varphi(x)(H)=H$ i.e. $Hx=H$ that is, $x\in H$, and conversely, $h\in H$ implies $\phi(h)=\mathrm{id}$ as $Hh=H$.
A: The map $\phi\colon N_G(H)\to\operatorname{Aut}_G(G/H)$ defined by $\phi(g)(g'H)=g'g^{-1}H$ is well defined as $g\in N_G(H)$, and
$$
\phi(g)(g'H)=g'g^{-1}H=g'\phi(g)(H)
$$
and
$$
\phi(gk)(g'H)=g'(gk)^{-1}H=g'k^{-1}g^{-1}H=\phi(g)(g'k^{-1}H)=\phi(g)\circ\phi(k)(g'H).
$$
This implies $\phi(g)\phi(g^{-1})=\phi(e)=1_{\operatorname{Aut}_G(G/H))}$. So $\phi$ is a group homomorphism, and each $\phi(g)$ is an $G$-invariant automorphism of $G/H$. 
It is surjective, because a $G$-invariant automorphism is determined by its image on $H$. So if $f(H)=kH$, then $f=\phi(k^{-1})$, as $\phi(k^{-1})(H)=kH=f(H)$. Also, $\phi(g)$ is the identity if it fixes $H$, that is, if $g^{-1}H=H$, i.e., iff $g\in H$. Thus $\ker(\phi)=H$. So by the first isomorphism, $\phi$ descends to an isomorphism $N_G(H)/H\simeq\operatorname{Aut}_G(G/H)$. 
