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Let $G$ be a group and $X$ a set equipped with a transitive right $G$-action. Further, let $c: X\to X$ be a $G$-equivariant map. Is it true that $\text{Stab}(x) = \text{Stab}(c(x))$ for all $x\in X$?

This doesn't seem to be an interesting mathoverflow question on its own, but the reason I ask is the following: In Hovey's book on Model Categories Hovey proves some sort of five lemma for pointed model categories (Thm.6.5.3). In the end of the proof, he seems to conclude from $\alpha\text{Stab}(x)\alpha^{-1}\subset\text{Stab}(x)$ that equality holds; I don't understand how this can be done. The above statement comes from a try to fix this gap.

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have you read the proof in the version which is available on google books? there he seems to eleborate the argument. –  Martin Brandenburg Jan 16 '10 at 13:34
When hovey states his claim, does he say it's true for any $\alpha$? Because then it's a very easy exercise to verify that it makes sense, since if we have a containment relation between those two, we can also see that the inverse conjugate contains Stab(x), but since it's true for $\alpha$ and $\alpha^{-1}$, we have equality. –  Harry Gindi Jan 16 '10 at 14:13
@Martin: The point is that I do not understand why $\alpha^{-1}$ should stabilize $\text{Stab}(x)$. Can you explain it to me? @Harry: Sorry, I should've said that $\alpha$ is some fixed element. Of course, if the relation would hold for all $\alpha$, then it'd be easy. –  Hanno Becker Jan 16 '10 at 14:41

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up vote 7 down vote accepted

That's essentially the same as this question. So the answer is negative: if $X=H\backslash G$, the stabilizer of $Ha\in X$ is $a^{-1}Ha$; if $b$ is any element such that $bHb^{-1}\subset H$, the map $$Ha\mapsto Hba$$ is $G$-equivariant; if $bHb^{-1}\ne H$, we get a counterexample.

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Thank you very much, that's a nice counterexample! But this also means that Hovey's proof is still unclear to me. –  Hanno Becker Jan 16 '10 at 14:44

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