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Let $H, K$ be conjugate subgroups of $G$. Then we can regard $G/H, G/K$ as G-sets defining $g'(gH) := (g'g)H$ and similarly for $K$.

Two G-sets $X, Y$ are said to be isomorphic if there exists some bijection $\phi: X \rightarrow Y$ such that

$$\phi(gx) = g\phi(x) \; \forall x \in X, g \in G$$

My intuition seems to be that $\phi(H) = K$ and so $\phi(gH) = gK$, but I'm having trouble trying to show $\phi$ is well defined, since $\phi(ghH) = ghK =_{?} gK = \phi(gH)$ does not look right.

I've found this question here in this site, but the only answer is rather brief and I don't fully understand what is going on, as seems to be the case for the OP. Anyway, it was never restated or further detailed, so I'd really appreciate if someone would care to give some insight into it.

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I will try to explain better the answer which appears in the post you mentioned.

To define a $G$-map $\phi \colon G/H \to G/K$ is equivalent to fixing some $g \in G$ such that $\phi(1H)=gK$ (note that $g$ does not need to be $1$). Indeed, then we have automatically $\phi(xH)=x\phi(1H)=xgK$. But $\phi$ is well-defined iff for all $h\in H$ we have $hgK=\phi(hH)=\phi(1H)=gK$, which is equivalent to $g^{-1}hgK = K$ for all $h$, which in turn is equivalent to $g^{-1}Hg \subseteq K$.

Now we show that the other inclusion is instead equivalent to injectivity of $\phi$. Note that $xgK=gK$ is equivalent to $x \in gKg^{-1}$. Now, injectivity is equivalent to requiring that all such $x$ are in $H$, i.e. $gKg^{-1}\subseteq H$.

Surjectivity is automatic for $\phi$, because if we want to find $x$ such that $\phi(xH)=xgK=qK$, we just need to set $x:=qg^{-1}$. Hence we conclude:

$\phi\colon G/H \to G/K, xH \mapsto xgK$ is an isomorphism of $G$-sets iff $H$ and $K$ are $g$-conjugated.

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One version of the orbit-stabilizer theorem is that $\mathrm{Orb}(x)\cong G/\mathrm{Stab}(x)$ as $G$-sets, via the correspondence $\color{Green}{g}\color{Blue}{x}\leftrightarrow \color{Green}{g}\color{Blue}{\mathrm{Stab}(x)}$. To prove this, one checks the fibers of $G\to\mathrm{Orb}(x)$ given by $g\mapsto gx$ are precisely the cosets of $\mathrm{Stab}(x)$. More generally, in the context of Lie groups, we can say that $\mathrm{Stab}(x)\to G\to\mathrm{Orb}(x)$ is a fiber bundle (fibers being the cosets).

The $G$-set $X=G/H$ is transitive, so by orbit-stabilizer we have

$$G/H=\mathrm{Orb}(\sigma H)\cong G/\mathrm{Stab}(\sigma H)=G/\sigma H\sigma^{-1}$$

for any $\sigma\in G$. That is, $G/H\cong G/\sigma H\sigma^{-1}$ as $G$-sets. Following along above, we have

$$ gH= (\color{Green}{g\sigma^{-1}})(\color{Blue}{\sigma H})\leftrightarrow (\color{Green}{g\sigma^{-1}})(\color{Blue}{\sigma H\sigma^{-1}}). $$

So the isomorphism $G/H\cong G/\sigma H\sigma^{-1}$ is given by $gH\mapsto (g\sigma^{-1})(\sigma H\sigma^{-1})$. To better see that it's well-defined, observe that this is $gH\mapsto gH\sigma^{-1}$ - that is, it's just right multiplication by $\sigma^{-1}$: this commutes with the left action of $G$ and has an obvious inverse (right multiplication by $\sigma$).

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Second isomorphism theorem: Let G be a group and HK < G. If H Δ G, then K/HnK ≅ HK / H

Define $\phi : K \to H k/H$ by $\phi(a) = aH$. I state that $\phi$ is well-defined homomorphism and endomorphism at the same time, this means that $\operatorname{Ker}f = HnK$

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