Isomorphism between the actions of $G$ on $G/H$ and $G/gHg^{-1}$? I'm doing Aluffi's Chapter 0, and Exercise 9.13 asks:

Prove that for all subgroups $H$ of a group $G$ and for all $g\in G$, $G/H$ and
  $G/(gHg^{-1})$ (endowed with the action of G by left-multiplication) are isomorphic in G-Set.

My plan is to first find a bijection between $G/H$ and $G/(gHg^{-1})$, then prove that the map commutes with the action.
The first map I thought of was just $\varphi:G/H \longrightarrow G/(gHg^{-1})$ defined by $aH \mapsto a(gHg^{-1})$, but the problem is that this is not well defined, since we can have $aH = a'H$ but $a(gHg^{-1}) \neq a'(gHg^{-1})$. 
What would be well-defined is $aH \mapsto g(aH)g^{-1}$. This works as a bijection between $G/H$ and $G/(gHg^{-1})$, but the problem is that is does not commute with the action of $G$. If the action wasn't strictly left-multiplication, then I could make the map work (by making it conjugate instead), but Aluffi is specifying that the action must be left multiplication. 
So now I don't really know what to do. On one hand, my gut feeling tells me that left-multiplication is just not "compatible" with conjugation, and maybe there's no way to make $\varphi$ commute with the action of $G$. On the other hand, Aluffi's exercises are pretty well done, and I might be just missing something?
 A: We want a bijection that is well-defined. If $xH=yH$, then $y^{-1}x\in H$, hence $gy^{-1}xg^{-1}\in gHg^{-1}$, hence $yg^{-1}(gHg^{-1}) = xg^{-1}(gHg^{-1})$. 
So, how about mapping the coset $aH$ to the coset $ag^{-1}(gHg^{-1})$?
Why that and not try something like “hence $gy^{-1}g^{-1}gxg^{-1}\in gHg^{-1}$, hence $gxg^{-1}(gHg^{-1}) = gyg^{-1}(gHg^{-1})$; so let’s define the image of $aH$ to be $gag^{-1}(gHg^{-1})$”, which was your second attempt? Because that extra $g$ on the left interferes with the action of $G$, whereas mapping $aH$ to $ag^{—1}(gHg^{-1})$ won’t run into that problem.
Alternatively, you can recognize the map $aH\mapsto ag^{-1}(gHg^{-1})$ as the result of your second map followed by “multiply by $g^{-1}$ on the left”. Essentially, it partially does the conjugation you want to do, but codes it into the bijection instead of into the action.
A: *

*Consider $G$ acting on the left cosets of $g^{-1}Hg\leq G$ by the left multiplication

*Note that this action is transitive: given $a(g^{-1}Hg)$, $b(g^{-1}Hg)$, we have $a(g^{-1}Hg)=ab^{-1}(b(g^{-1}Hg))$

*$H=Stab_G(g(g^{-1}Hg))=Stab_G(Hg)$ since $\forall(h\in H):hHg=Hg$ and $xHg=Hg\Rightarrow xH=H\Rightarrow x\in H$

*Apply Proposition 9.9 from the Aluffi's book


P.S. I'm also studying Algebra: Chapter 0 so we can work together 
