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It is not true : for instance, $X$ and $X'$ might be two non-isomorphic $G$-torsors on the same $S$-scheme. S$-scheme$T$. Here is a precise example. Consider$S=Spec(\mathbb{R})$S=T=Spec(\mathbb{R})$ and $G=\mathbb{Z}/2\mathbb{Z}$, let $X=Spec(\mathbb{C})$ and $X'=Spec(\mathbb{R})\cup Spec(\mathbb{R})$ viewed as $S$-schemes, and let $G$ act on $X$ and $X'$ respectively by complex conjugation and by exchanging the connected components. Both actions admit $Spec(\mathbb{R})$ as a geometric quotient. But an isomorphism between these quotients obviously doesn't lift to an equivariant morphism between $X$ and $X'$.

Here is another example : take $S=Spec(\mathbb{C})$, $G=\mathbb{G}_m$, $T=\mathbb{P}^1_{\mathbb{C}}$. Let $X$ (resp. $X'$) be the total space of the line bundle $\mathcal{O}$ (resp. $\mathcal{O}(1)$) on $T$ minus the zero-section, with the natural $G$-action. Both actions admit $T$ as a geometric quotient, but there is no equivariant morphism between $X$ and $X'$ lifting identity.

3 added 404 characters in body

It is not true : for instance, $X$ and $X'$ might be two non-isomorphic $G$-torsors on the same $S$. S$-scheme. Here is a precise example. Consider$S=Spec(\mathbb{R})$and$G=\mathbb{Z}/2\mathbb{Z}$, let$X=Spec(\mathbb{C})$and$X'=Spec(\mathbb{R})\cup Spec(\mathbb{R})$viewed as$S$-schemes, and let$G$act on$X$and$X'$respectively by complex conjugation and by exchanging the connected components. Both actions admit$Spec(\mathbb{R})$as a geometric quotient. But an isomorphism between these quotients obviously doesn't lift to an equivariant morphism between$X$and$X'$. Here is another example : take$S=Spec(\mathbb{C})$,$G=\mathbb{G}_m$, $T=\mathbb{P}^1_{\mathbb{C}}$. Let$X$(resp.$X'$) be the total space of the line bundle$\mathcal{O}$(resp.$\mathcal{O}(1)$) on$T$minus the zero-section, with the natural$G$-action. Both actions admit$T$as a geometric quotient, but there is no equivariant morphism between$X$and$X'$lifting identity. 2 added 98 characters in body It is not true . For : for instance, consider$S=Spec(\mathbb{R})$,$G=\mathbb{Z}/2\mathbb{Z}$, and let$X$and$X'$might be two non-isomorphic$Spec(\mathbb{C})$G$-torsors on $S$.

Here is a precise example. Consider $S=Spec(\mathbb{R})$ and $Spec(\mathbb{R})\cup G=\mathbb{Z}/2\mathbb{Z}$, let $X=Spec(\mathbb{C})$ and $X'=Spec(\mathbb{R})\cup Spec(\mathbb{R})$ viewed as $S$-schemes, and let $G$ act on $X$ and $X'$ respectively by complex conjugation and by exchanging the factorsconnected components. Both actions admit $Spec(\mathbb{R})$ as a geometric quotient. But an isomorphism between these quotients obviously doesn't lift to an equivariant morphism between $X$ and $X'$.

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