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That is not true. The basic issue is that for every coherent sheaf $\mathcal{F}$ on $X$ that is "intrinsic", and thus admits a linearization of the automorphism group, every cohomology group of $\mathcal{F}$ gives a linear representation of the automorphism group. So for every form of $X$, not only do you get a form of the automorphism group, you get a collection of linear representations for every intrinstic coherent sheaf $\mathcal{F}$. It can easily happen that some form of the automorphism group does not have "enough" linear representations over $k$ to arise from a form of $X$.

Here is a concrete counterexample. Let $(a,b,c,d)$ be positive integers with $a\neq b$, with $c\neq d$, and with the sums $m=a+b$ and $n=c+d$ distinct. Let $A$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[s,t]$. Let $C$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[u,v]$. Let $P$ be a copy of $\mathbb{P}^3$ with homogeneous coordinates $[x_0,x_1,x_2,x_3]$. Let $Q$ be a copy of $\mathbb{P}^3$ with homogeneous coordinates $[y_0,y_1,y_2,y_3]$. Denote by $f_{a,b}$, resp. $g_{c,d}$, the closed immersions, $$f_{a,b}:A \to P, \ \ [s,t] \mapsto [s^m,s^{m-1}t,s^at^b,st^{m-1},t^m],$$ $$g_{c,d}:C \to Q, \ \ [u,v] \mapsto [u^n,u^{n-1}v,u^cv^d,uv^{n-1},v^n].$$ Let $R$ be a copy of $\mathbb{P}^{15}$, and denote by $h$ the Segre embedding, $$h:P\times Q \to R,\ \ ([x_i],[y_j]) \mapsto [x_iy_j]_{0\leq i,j\leq 3}.$$ Define $Y$ to be the blowing up of $R$ along the image of $P\times Q$, and define $X$ to be the blowing up of $Y$ along the inverse image of $A\times C$.

The automorphism group of $X$ equals the subgroup of the automorphism group of $Y$ that preserves the inverse image of $A\times C$ as a closed subscheme (rather than "pointwise"). The automorphism group of $Y$ is a semidirect product of $\text{Aut}(P)\times \text{Aut}(Q)$ by a cyclic group $\mathfrak{S}_2$ permuting the factors. The automorphism group of $X$ is the subgroup of this group that preserves $A\times C$ as a closed subscheme of $P\times Q$. Because $m$ does not equal $n$, the cyclic group $\mathfrak{S}_2$ does not preserve $A\times C$. Because $a\neq b$, resp. $c\neq d$, the automorphism group of $A$ in $P$, resp. the automorphism group of $C$ in $Q$, is a copy of the multiplicative group $\mathbb{G}_m$. Thus, the automorphism group of $X$ is isomorphic to $\mathbb{G}_m\times \mathbb{G}_m$. There is a linear representation of this automorphism group on the $k$-vector space of global sections on $X$ of the pullback of $\mathcal{O}(1)$ from $R$. This representation is a direct sum of copies of the characters with weights $(i,j)$ for $$i\in\{ 0,1,a,m-1,m\},\ \ j\in \{ 0,1,c,n-1,n\}.$$ Similarly, the representation on global sections of the pullback of $\mathcal{O}(16) \cong \omega_{R/k}^\vee$ has weights $(i,j)$ ranging from $(0,0)$ to $(16m,16n)$.

The automorphism group of the group scheme $\mathbb{G}_m\times \mathbb{G}_m$ is quite large. For instance, for every separable, degree $2$-field extension $L/k$, the restriction $\text{Res}_{L/k}(\mathbb{G}_{m,L})$ is a group $k$-scheme that is a form of $\mathbb{G}_m\times \mathbb{G}_m$. Since $m$ does not equal $n$, there is no realization of this form arising from a form of $X$: if there were, then the linear representation on the global sections of the twist of the pullback of $\omega_{X/k}^\vee$ would have weights $(i,j)$ that were symmetric under permutation of the two factors of $\mathbb{G}_m\times \mathbb{G}_m$.

There should be similar examples obtained from blowing up finite sets in projective space, but this requires explicit computation of the stabilizer group of the finite point set.

That is not true. The basic issue is that for every coherent sheaf $\mathcal{F}$ on $X$ that is "intrinsic", and thus admits a linearization of the automorphism group, every cohomology group of $\mathcal{F}$ gives a linear representation of the automorphism group. So for every form of $X$, not only do you get a form of the automorphism group, you get a collection of linear representations for every intrinstic coherent sheaf $\mathcal{F}$. It can easily happen that some form of the automorphism group does not have "enough" linear representations over $k$ to arise from a form of $X$.

Here is a concrete counterexample. Let $(a,b,c,d)$ be positive integers with $a\neq b$, with $c\neq d$, and with the sums $m=a+b$ and $n=c+d$ distinct. Let $A$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[s,t]$. Let $C$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[u,v]$. Let $P$ be a copy of $\mathbb{P}^4$ with homogeneous coordinates $[x_0,x_1,x_2,x_3,x_4]$. Let $Q$ be a copy of $\mathbb{P}^4$ with homogeneous coordinates $[y_0,y_1,y_2,y_3,y_4]$. Denote by $f_{a,b}$, resp. $g_{c,d}$, the closed immersions, $$f_{a,b}:A \to P, \ \ [s,t] \mapsto [s^m,s^{m-1}t,s^at^b,st^{m-1},t^m],$$ $$g_{c,d}:C \to Q, \ \ [u,v] \mapsto [u^n,u^{n-1}v,u^cv^d,uv^{n-1},v^n].$$ Let $R$ be a copy of $\mathbb{P}^{24}$, and denote by $h$ the Segre embedding, $$h:P\times Q \to R,\ \ ([x_i],[y_j]) \mapsto [x_iy_j]_{0\leq i,j\leq 4}.$$ Define $Y$ to be the blowing up of $R$ along the image of $P\times Q$, and define $X$ to be the blowing up of $Y$ along the inverse image of $A\times C$.

The automorphism group of $X$ equals the subgroup of the automorphism group of $Y$ that preserves the inverse image of $A\times C$ as a closed subscheme (rather than "pointwise"). The automorphism group of $Y$ is a semidirect product of $\text{Aut}(P)\times \text{Aut}(Q)$ by a cyclic group $\mathfrak{S}_2$ permuting the factors. The automorphism group of $X$ is the subgroup of this group that preserves $A\times C$ as a closed subscheme of $P\times Q$. Because $m$ does not equal $n$, the cyclic group $\mathfrak{S}_2$ does not preserve $A\times C$. Because $a\neq b$, resp. $c\neq d$, the automorphism group of $A$ in $P$, resp. the automorphism group of $C$ in $Q$, is a copy of the multiplicative group $\mathbb{G}_m$. Thus, the automorphism group of $X$ is isomorphic to $\mathbb{G}_m\times \mathbb{G}_m$. There is a linear representation of this automorphism group on the $k$-vector space of global sections on $X$ of the pullback of $\mathcal{O}(1)$ from $R$. This representation is a direct sum of copies of the characters with weights $(i,j)$ for $$i\in\{ 0,1,a,m-1,m\},\ \ j\in \{ 0,1,c,n-1,n\}.$$ Similarly, the representation on global sections of the pullback of $\mathcal{O}(16) \cong \omega_{R/k}^\vee$ has weights $(i,j)$ ranging from $(0,0)$ to $(16m,16n)$.

The automorphism group of the group scheme $\mathbb{G}_m\times \mathbb{G}_m$ is quite large. For instance, for every separable, degree $2$-field extension $L/k$, the restriction $\text{Res}_{L/k}(\mathbb{G}_{m,L})$ is a group $k$-scheme that is a form of $\mathbb{G}_m\times \mathbb{G}_m$. Since $m$ does not equal $n$, there is no realization of this form arising from a form of $X$: if there were, then the linear representation on the global sections of the twist of the pullback of $\omega_{X/k}^\vee$ would have weights $(i,j)$ that were symmetric under permutation of the two factors of $\mathbb{G}_m\times \mathbb{G}_m$.

There should be similar examples obtained from blowing up finite sets in projective space, but this requires explicit computation of the stabilizer group of the finite point set.

That is not true. The basic issue is that for every coherent sheaf $\mathcal{F}$ on $X$ that is "intrinstic", and thus admits a linearization of the automorphism group, every cohomology group of $\mathcal{F}$ gives a linear representation of the automorphism group. So for every form of $X$, not only do you get a form of the automorphism group, you get a collection of linear representations for every intrinstic coherent sheaf $\mathcal{F}$. It can easily happen that some form of the automorphism group does not have "enough" linear representations over $k$ to arise from a form of $X$.

Here is a concrete counterexample. Let $(a,b,c,d)$ be positive integers with $a\neq b$, with $c\neq d$, and with the sums $m=a+b$ and $n=c+d$ distinct. Let $A$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[s,t]$. Let $C$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[u,v]$. Let $P$ be a copy of $\mathbb{P}^3$ with homogeneous coordinates $[x_0,x_1,x_2,x_3]$. Let $Q$ be a copy of $\mathbb{P}^3$ with homogeneous coordinates $[y_0,y_1,y_2,y_3]$. Denote by $f_{a,b}$, resp. $g_{c,d}$, the closed immersions, $$f_{a,b}:A \to P, \ \ [s,t] \mapsto [s^m,s^{m-1}t,s^at^b,st^{m-1},t^m],$$ $$g_{c,d}:C \to Q, \ \ [u,v] \mapsto [u^n,u^{n-1}v,u^cv^d,uv^{n-1},v^n].$$ Let $R$ be a copy of $\mathbb{P}^{15}$, and denote by $h$ the Segre embedding, $$h:P\times Q \to R,\ \ ([x_i],[y_j]) \mapsto [x_iy_j]_{0\leq i,j\leq 3}.$$ Define $Y$ to be the blowing up of $R$ along the image of $P\times Q$, and define $X$ to be the blowing up of $Y$ along the inverse image of $A\times C$.

The automorphism group of $X$ equals the subgroup of the automorphism group of $Y$ that preserves the inverse image of $A\times C$ as a closed subscheme (rather than "pointwise"). The automorphism group of $Y$ is a semidirect product of $\textbf{Aut}(P)\times \textbf{Aut}(Q)$ by a cyclic group $\mathfrak{S}_2$ permuting the factors. The automorphism group of $X$ is the subgroup of this group that preserves $A\times C$ as a closed subscheme of $P\times Q$. Because $m$ does not equal $n$, the cyclic group $\mathfrak{S}_2$ does not preserve $A\times C$. Because $a\neq b$, resp. $c\neq d$, the automorphism group of $A$ in $P$, resp. the automorphism group of $C$ in $Q$, is a copy of the mutliplicative group $\textbf{G}_m$. Thus, the automorphism group of $X$ is isomorphic to $\mathbb{G}_m\times \mathbb{G}_m$. There is a linear representation of this automorphism group on the $k$-vector space of global sections on $X$ of the pullback of $\mathcal{O}(1)$ from $R$. This representation is a direct sum of copies of the characters with weights $(i,j)$ for $$i\in\{ 0,1,a,m-1,m\},\ \ j\in \{ 0,1,c,n-1,n\}.$$ Similarly, the representation on global sections of the pullback of $\mathcal{O}(16) \cong \omega_{R/k}^\vee$ has weights $(i,j)$ ranging from $(0,0)$ to $(16m,16n)$.

The automorphism group of the group scheme $\mathbb{G}_m\times \mathbb{G}_m$ is quite large. For instance, for every separable, degree $2$-field extension $L/k$, the restriction $\text{Res}_{L/k}(\mathbb{G}_{m,L})$ is a group $k$-scheme that is a form of $\mathbb{G}_m\times \mathbb{G}_m$. Since $m$ does not equal $n$, there is no realization of this form arising from a form of $X$: if there were, then the linear representation on the global sections of the twist of the pullback of $\omega_{X/k}^\vee$ would have weights $(i,j)$ that were symmetric under permutation of the two factors of $\mathbb{G}_m\times \mathbb{G}_m$.

There should be similar examples obtained from blowing up finite sets in projective space, but this requires explicit computation of the stabilizer group of the finite point set.

That is not true. The basic issue is that for every coherent sheaf $\mathcal{F}$ on $X$ that is "intrinsic", and thus admits a linearization of the automorphism group, every cohomology group of $\mathcal{F}$ gives a linear representation of the automorphism group. So for every form of $X$, not only do you get a form of the automorphism group, you get a collection of linear representations for every intrinstic coherent sheaf $\mathcal{F}$. It can easily happen that some form of the automorphism group does not have "enough" linear representations over $k$ to arise from a form of $X$.

Here is a concrete counterexample. Let $(a,b,c,d)$ be positive integers with $a\neq b$, with $c\neq d$, and with the sums $m=a+b$ and $n=c+d$ distinct. Let $A$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[s,t]$. Let $C$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[u,v]$. Let $P$ be a copy of $\mathbb{P}^3$ with homogeneous coordinates $[x_0,x_1,x_2,x_3]$. Let $Q$ be a copy of $\mathbb{P}^3$ with homogeneous coordinates $[y_0,y_1,y_2,y_3]$. Denote by $f_{a,b}$, resp. $g_{c,d}$, the closed immersions, $$f_{a,b}:A \to P, \ \ [s,t] \mapsto [s^m,s^{m-1}t,s^at^b,st^{m-1},t^m],$$ $$g_{c,d}:C \to Q, \ \ [u,v] \mapsto [u^n,u^{n-1}v,u^cv^d,uv^{n-1},v^n].$$ Let $R$ be a copy of $\mathbb{P}^{15}$, and denote by $h$ the Segre embedding, $$h:P\times Q \to R,\ \ ([x_i],[y_j]) \mapsto [x_iy_j]_{0\leq i,j\leq 3}.$$ Define $Y$ to be the blowing up of $R$ along the image of $P\times Q$, and define $X$ to be the blowing up of $Y$ along the inverse image of $A\times C$.

The automorphism group of $X$ equals the subgroup of the automorphism group of $Y$ that preserves the inverse image of $A\times C$ as a closed subscheme (rather than "pointwise"). The automorphism group of $Y$ is a semidirect product of $\text{Aut}(P)\times \text{Aut}(Q)$ by a cyclic group $\mathfrak{S}_2$ permuting the factors. The automorphism group of $X$ is the subgroup of this group that preserves $A\times C$ as a closed subscheme of $P\times Q$. Because $m$ does not equal $n$, the cyclic group $\mathfrak{S}_2$ does not preserve $A\times C$. Because $a\neq b$, resp. $c\neq d$, the automorphism group of $A$ in $P$, resp. the automorphism group of $C$ in $Q$, is a copy of the multiplicative group $\mathbb{G}_m$. Thus, the automorphism group of $X$ is isomorphic to $\mathbb{G}_m\times \mathbb{G}_m$. There is a linear representation of this automorphism group on the $k$-vector space of global sections on $X$ of the pullback of $\mathcal{O}(1)$ from $R$. This representation is a direct sum of copies of the characters with weights $(i,j)$ for $$i\in\{ 0,1,a,m-1,m\},\ \ j\in \{ 0,1,c,n-1,n\}.$$ Similarly, the representation on global sections of the pullback of $\mathcal{O}(16) \cong \omega_{R/k}^\vee$ has weights $(i,j)$ ranging from $(0,0)$ to $(16m,16n)$.

The automorphism group of the group scheme $\mathbb{G}_m\times \mathbb{G}_m$ is quite large. For instance, for every separable, degree $2$-field extension $L/k$, the restriction $\text{Res}_{L/k}(\mathbb{G}_{m,L})$ is a group $k$-scheme that is a form of $\mathbb{G}_m\times \mathbb{G}_m$. Since $m$ does not equal $n$, there is no realization of this form arising from a form of $X$: if there were, then the linear representation on the global sections of the twist of the pullback of $\omega_{X/k}^\vee$ would have weights $(i,j)$ that were symmetric under permutation of the two factors of $\mathbb{G}_m\times \mathbb{G}_m$.

There should be similar examples obtained from blowing up finite sets in projective space, but this requires explicit computation of the stabilizer group of the finite point set.

Here is a concrete counterexample. Let $(a,b,c,d)$ be positive integers with $a\neq b$, with $c\neq d$, and with the sums $m=a+b$ and $n=c+d$ distinct. Let $A$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[s,t]$. Let $C$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[u,v]$. Let $P$ be a copy of $\mathbb{P}^3$ with homogeneous coordinates $[x_0,x_1,x_2,x_3]$. Let $Q$ be a copy of $\mathbb{P}^3$ with homogeneous coordinates $[y_0,y_1,y_2,y_3]$. Denote by $f_{a,b}$, resp. $g_{c,d}$, the closed immersions, $$f_{a,b}:A \to P, \ \ [s,t] \mapsto [s^m,s^{m-1}t,s^at^b,st^{m-1},t^m],$$ $$g_{c,d}:C \to Q, \ \ [u,v] \mapsto [u^n,u^{n-1}v,u^cv^d,uv^{n-1},v^n].$$ Let $R$ be a copy of $\mathbb{P}^{15}$, and denote by $h$ the Segre embedding, $$h:P\times Q \to R,\ \ ([x_i],[y_j]) \mapsto [x_iy_j]_{0\leq i,j\leq 3}.$$ Define $Y$ to be the blowing up of $R$ along the image of $P\times Q$, and define $X$ to be the blowing up of $Y$ along the inverse image of $A\times C$.

The automorphism group of $X$ equals the subgroup of the automorphism group of $Y$ that preserves the inverse image of $A\times C$ as a closed subscheme (rather than "pointwise"). The automorphism group of $Y$ is a semidirect product of $\textbf{Aut}(P)\times \textbf{Aut}(Q)$ by a cyclic group $\mathfrak{S}_2$ permuting the factors. The automorphism group of $X$ is the subgroup of this group that preserves $A\times C$ as a closed subscheme of $P\times Q$. Because $m$ does not equal $n$, the cyclic group $\mathfrak{S}_2$ does not preserve $A\times C$. Because $a\neq b$, resp. $c\neq d$, the automorphism group of $A$ in $P$, resp. the automorphism group of $C$ in $Q$, is a copy of the mutliplicative group $\textbf{G}_m$. Thus, the automorphism group of $X$ is isomorphic to $\mathbb{G}_m\times \mathbb{G}_m$. There is a linear representation of this automorphism group on the $k$-vector space of global sections on $X$ of the pullback of $\mathcal{O}(1)$ from $R$. This representation is a direct sum of copies of the characters with weights $(i,j)$ for $$i\in\{ 0,1,a,m-1,m\},\ \ j\in \{ 0,1,c,n-1,n\}.$$ Similarly, the representation on global sections of the pullback of $\mathcal{O}(16) \cong \omega_{R/k}^\vee$ has weights $(i,j)$ ranging from $(0,0)$ to $(16m,16n)$.

The automorphism group of the group scheme $\mathbb{G}_m\times \mathbb{G}_m$ is quite large. For instance, for every separable, degree $2$-field extension $L/k$, the restriction $\text{Res}_{L/k}(\mathbb{G}_{m,L})$ is a group $k$-scheme that is a form of $\mathbb{G}_m\times \mathbb{G}_m$. Since $m$ does not equal $n$, there is no realization of this form arising from a form of $X$: if there were, then the linear representation on the global sections of the twist of the pullback of $\omega_{X/k}^\vee$ would have weights $(i,j)$ that were symmetric under permutation of the two factors of $\mathbb{G}_m\times \mathbb{G}_m$.

There should be similar examples obtained from blowing up finite sets in projective space, but this requires explicit computation of the stabilizer group of the finite point set.

That is not true. The basic issue is that for every coherent sheaf $\mathcal{F}$ on $X$ that is "intrinstic", and thus admits a linearization of the automorphism group, every cohomology group of $\mathcal{F}$ gives a linear representation of the automorphism group. So for every form of $X$, not only do you get a form of the automorphism group, you get a collection of linear representations for every intrinstic coherent sheaf $\mathcal{F}$. It can easily happen that some form of the automorphism group does not have "enough" linear representations over $k$ to arise from a form of $X$.

Here is a concrete counterexample. Let $(a,b,c,d)$ be positive integers with $a\neq b$, with $c\neq d$, and with the sums $m=a+b$ and $n=c+d$ distinct. Let $A$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[s,t]$. Let $C$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[u,v]$. Let $P$ be a copy of $\mathbb{P}^3$ with homogeneous coordinates $[x_0,x_1,x_2,x_3]$. Let $Q$ be a copy of $\mathbb{P}^3$ with homogeneous coordinates $[y_0,y_1,y_2,y_3]$. Denote by $f_{a,b}$, resp. $g_{c,d}$, the closed immersions, $$f_{a,b}:A \to P, \ \ [s,t] \mapsto [s^m,s^{m-1}t,s^at^b,st^{m-1},t^m],$$ $$g_{c,d}:C \to Q, \ \ [u,v] \mapsto [u^n,u^{n-1}v,u^cv^d,uv^{n-1},v^n].$$ Let $R$ be a copy of $\mathbb{P}^{15}$, and denote by $h$ the Segre embedding, $$h:P\times Q \to R,\ \ ([x_i],[y_j]) \mapsto [x_iy_j]_{0\leq i,j\leq 3}.$$ Define $Y$ to be the blowing up of $R$ along the image of $P\times Q$, and define $X$ to be the blowing up of $Y$ along the inverse image of $A\times C$.

The automorphism group of $X$ equals the subgroup of the automorphism group of $Y$ that preserves the inverse image of $A\times C$ as a closed subscheme (rather than "pointwise"). The automorphism group of $Y$ is a semidirect product of $\textbf{Aut}(P)\times \textbf{Aut}(Q)$ by a cyclic group $\mathfrak{S}_2$ permuting the factors. The automorphism group of $X$ is the subgroup of this group that preserves $A\times C$ as a closed subscheme of $P\times Q$. Because $m$ does not equal $n$, the cyclic group $\mathfrak{S}_2$ does not preserve $A\times C$. Because $a\neq b$, resp. $c\neq d$, the automorphism group of $A$ in $P$, resp. the automorphism group of $C$ in $Q$, is a copy of the mutliplicative group $\textbf{G}_m$. Thus, the automorphism group of $X$ is isomorphic to $\mathbb{G}_m\times \mathbb{G}_m$. There is a linear representation of this automorphism group on the $k$-vector space of global sections on $X$ of the pullback of $\mathcal{O}(1)$ from $R$. This representation is a direct sum of copies of the characters with weights $(i,j)$ for $$i\in\{ 0,1,a,m-1,m\},\ \ j\in \{ 0,1,c,n-1,n\}.$$ Similarly, the representation on global sections of the pullback of $\mathcal{O}(16) \cong \omega_{R/k}^\vee$ has weights $(i,j)$ ranging from $(0,0)$ to $(16m,16n)$.

The automorphism group of the group scheme $\mathbb{G}_m\times \mathbb{G}_m$ is quite large. For instance, for every separable, degree $2$-field extension $L/k$, the restriction $\text{Res}_{L/k}(\mathbb{G}_{m,L})$ is a group $k$-scheme that is a form of $\mathbb{G}_m\times \mathbb{G}_m$. Since $m$ does not equal $n$, there is no realization of this form arising from a form of $X$: if there were, then the linear representation on the global sections of the twist of the pullback of $\omega_{X/k}^\vee$ would have weights $(i,j)$ that were symmetric under permutation of the two factors of $\mathbb{G}_m\times \mathbb{G}_m$.

There should be similar examples obtained from blowing up finite sets in projective space, but this requires explicit computation of the stabilizer group of the finite point set.