This is false whenever $G \ne 1$ is either a split torus or connected semisimple and absolutely simple with $Y \ne X$. By passing to generic fiber over $X$, it suffices to show that if $K'/K$ is a finite separable extension of degree $d > 1$ and $H\ne 1$ is a split $K$-torus or is connected semisimple and absolutely simple then ${\rm{R}}_{K'/K}(H_{K'})$ is not $K$-isomorphic to $H^d$. (It is also false in plenty of other situations, but is not worth the effort to address even more generally.)

If $H=T$ is a nontrivial split $K$-torus then the inclusion $T \hookrightarrow {\rm{R}}_{K'/K}(T_{K'})$ is easily seen to be the maximal $K$-split torus, yet the Weil restriction has even bigger dimension.

The semisimple and absolutely simple case is (as always) more interesting. If $\widetilde{H} \rightarrow H$ is the simply connected central cover, then the same holds for ${\rm{R}}_{K'/K}(\widetilde{H}_{K'}) \rightarrow {\rm{R}}_{K'/K}(H_{K'})$, so we may assume $H$ is simply connected.
Thus, by 6.21(ii) in Groupes Reductifs (IHES 27) (which is ultimately an application of the product structure of root data for split simple connected groups), every connected semisimple $K$-group that is simply connected has the form ${\rm{R}}_{F/K}(G)$ a unique pair $(F,G)$ up to $K$-isomorphism consisting of a nonzero finite etale $K$-algebra $F$ and a smooth affine $F$-group $G$ such that the fibers of $G$ over the factor fields of $F$ are connected semisimple, simply connected, and absolutely simple.

Since $H^d = {\rm{R}}_{F/K}(H_F)$ for $F = K^d$ and $H$ that is absolutely simple over $K$, it follows that the only way $H^d$ can be $K$-isomorphic to ${\rm{R}}_{K'/K}(H_{K'})$ is if $K' \simeq F = K^d$ as $K$-algebras, an absurdity since $d > 1$ and $K'$ is a field.

This is false whenever $G \ne 1$ is either a split torus or connected semisimple and absolutely simple with $Y \ne X$. By passing to generic fiber over $X$, it suffices to show that if $K'/K$ is a finite separable extension of degree $d > 1$ and $H\ne 1$ is a split $K$-torus or is connected semisimple and absolutely simple then $\operatorname R_{K'/K}(H_{K'})$ is not $K$-isomorphic to $H^d$. (It is also false in plenty of other situations, but is not worth the effort to address even more generally.)

If $H=T$ is a nontrivial split $K$-torus then the inclusion $T \hookrightarrow \operatorname R_{K'/K}(T_{K'})$ is easily seen to be the maximal $K$-split torus, yet the Weil restriction has even bigger dimension.

The semisimple and absolutely simple case is (as always) more interesting. If $\widetilde{H} \rightarrow H$ is the simply connected central cover, then the same holds for $\operatorname R_{K'/K}(\widetilde{H}_{K'}) \rightarrow \operatorname R_{K'/K}(H_{K'})$, so we may assume $H$ is simply connected.
Thus, by 6.21(ii) in Borel and Tits - Groupes Réductifs (IHES 27) (which is ultimately an application of the product structure of root data for split simple connected groups), every connected semisimple $K$-group that is simply connected has the form $\operatorname R_{F/K}(G)$ a unique pair $(F,G)$ up to $K$-isomorphism consisting of a nonzero finite étale $K$-algebra $F$ and a smooth affine $F$-group $G$ such that the fibers of $G$ over the factor fields of $F$ are connected semisimple, simply connected, and absolutely simple.

Since $H^d = \operatorname R_{F/K}(H_F)$ for $F = K^d$ and $H$ that is absolutely simple over $K$, it follows that the only way $H^d$ can be $K$-isomorphic to $\operatorname R_{K'/K}(H_{K'})$ is if $K' \simeq F = K^d$ as $K$-algebras, an absurdity since $d > 1$ and $K'$ is a field.

This is never true when $G$ is connected reductive and $Y \ne X$. (Of course it is sometimes true when $G$ is a vector group.) In general, if $K'/K$ is a finite separable extension of fields with $d = [K':K] > 1$ and $H$ is a connected reductive $K$-group (think of $K$ as the function field of $X$ in the question posed, and $H = G \otimes_k K$) then ${\rm{R}}_{K'/K}(H_{K'})$ is never $K$-isomorphic to $H^d$ as a $K$-group.

Indeed, consideration of maximal central torus and semisimple quotient by that reduces it to separate consideration of tori and semisimple groups, and the semisimple case reduces to the simply connected case. So, finally, one just needs to separately treat the cases when $H\ne 1 $ either is a torus or is connected semisimple and simply connected. By a theorem of Borel-Tits (6.21(ii) in Groupes Reductifs, IHES 27), in the latter case $H = {\rm{R}}_{F/K}(G)$ for a unique pair $(F,G)$ consisting of a finite etale $K$-algebra $F$ and smooth affine $F$-group $G$ with fibers over the factor fields of $F$ that are connected semisimple, absolutely simple, and simply connected. (The pair $(F,G)$ is unique up to unique $k$-Hence, $H_{K'} = {\rm{R}}_{F'/K'}(G')$ where $F' = F \otimes_K K'$ and $G' = G \otimes_F F'$, so ${\rm{R}}{K'/K}(H{K'}) =

This is false whenever $G \ne 1$ is either a split torus or connected semisimple and absolutely simple with $Y \ne X$. By passing to generic fiber over $X$, it suffices to show that if $K'/K$ is a finite separable extension of degree $d > 1$ and $H\ne 1$ is a split $K$-torus or is connected semisimple and absolutely simple then ${\rm{R}}_{K'/K}(H_{K'})$ is not $K$-isomorphic to $H^d$. (It is also false in plenty of other situations, but is not worth the effort to address even more generally.)

If $H=T$ is a nontrivial split $K$-torus then the inclusion $T \hookrightarrow {\rm{R}}_{K'/K}(T_{K'})$ is easily seen to be the maximal $K$-split torus, yet the Weil restriction has even bigger dimension.

The semisimple and absolutely simple case is (as always) more interesting. If $\widetilde{H} \rightarrow H$ is the simply connected central cover, then the same holds for ${\rm{R}}_{K'/K}(\widetilde{H}_{K'}) \rightarrow {\rm{R}}_{K'/K}(H_{K'})$, so we may assume $H$ is simply connected.
Thus, by 6.21(ii) in Groupes Reductifs (IHES 27) (which is ultimately an application of the product structure of root data for split simple connected groups), every connected semisimple $K$-group that is simply connected has the form ${\rm{R}}_{F/K}(G)$ a unique pair $(F,G)$ up to $K$-isomorphism consisting of a nonzero finite etale $K$-algebra $F$ and a smooth affine $F$-group $G$ such that the fibers of $G$ over the factor fields of $F$ are connected semisimple, simply connected, and absolutely simple.

Since $H^d = {\rm{R}}_{F/K}(H_F)$ for $F = K^d$ and $H$ that is absolutely simple over $K$, it follows that the only way $H^d$ can be $K$-isomorphic to ${\rm{R}}_{K'/K}(H_{K'})$ is if $K' \simeq F = K^d$ as $K$-algebras, an absurdity since $d > 1$ and $K'$ is a field.