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The assumption implies that the natural embedding induces an isomorphism $\Gamma(X,\mathscr O_X)\simeq \Gamma(\overline X,\mathscr O_{\overline X})$. Then this means that the complement of $X$ has at least codimension $2$.

In addition assume that $X$ is noetherian and $S_2$ (for instance normal).

In this case if $\Omega_{\overline X}$ is a reflexive sheaf, then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{\overline X})=\Gamma(X,\Omega_{X})$ is an isomorphism.

More generally, let $Z:=\overline X\setminus X$. If $\mathrm{depth}_Z\Omega_{\overline X}\geq 2$ then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{X})$ is an isomorphism. This certainly holds if $\Omega_{\overline X}$ is a reflexive sheaf, but obviously it could hold "by accident" even if one of the above conditions fail, so I am not claiming that these are necessary conditions, but at least they sure seem to provide a natural set of conditions under which the required map is an isomorphism.

Sketch that if $X$ is $S_2$, then a reflexive coherent sheaf $\mathscr F$ is also $S_2$: First observe that by the argument in this answer to another MO question we may assume that $X$ is affine and it is enough to prove that $H^i_x(X,\mathscr F)=0$ for $i=0,1$ for all $x\in X$ and it also follows that $\mathrm{depth}_Z\mathscr F\geq 2$ even if $Z$ is not contained in an affine piece of $X$. To do that write $\mathscr F^\vee$ as the quotient of a (locally) free sheaf ($X$ is affine!). Then $\mathscr F$ is a submodule of the dual of this locally free sheaf, let's call it $\mathscr E$, and the quotient $\mathscr E/\mathscr F$ is torsion-free. Therefore none of them have torsion and so $H^0_x(X,\mathscr F)=0$ and $H^1_x(X,\mathscr F)$ embeds into $H^1_x(X,\mathscr E)$. But the latter is $0$ by the assumption that $X$ is $S_2$.

EDIT 1 removed intro paragraph about the starting assumption.

EDIT 2 added "more generally" paragraph.

EDIT 3 added noetherian assumption. this is probably not necessary but without this one should possibly be more careful about the other conditions.

EDIT 4 Added Sketch above.

The assumption implies that the natural embedding induces an isomorphism $\Gamma(X,\mathscr O_X)\simeq \Gamma(\overline X,\mathscr O_{\overline X})$. Then this means that the complement of $X$ has at least codimension $2$.

In addition assume that $X$ is noetherian and $S_2$ (for instance normal).

In this case if $\Omega_{\overline X}$ is a reflexive sheaf, then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{\overline X})=\Gamma(X,\Omega_{X})$ is an isomorphism.

More generally, let $Z:=\overline X\setminus X$. If $\mathrm{depth}_Z\Omega_{\overline X}\geq 2$ then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{X})$ is an isomorphism. This certainly holds if $\Omega_{\overline X}$ is a reflexive sheaf, but obviously it could hold "by accident" even if one of the above conditions fail, so I am not claiming that these are necessary conditions, but at least they sure seem to provide a natural set of conditions under which the required map is an isomorphism.

Sketch that if $X$ is $S_2$, then a reflexive coherent sheaf $\mathscr F$ is also $S_2$: First observe that by the argument in this answer to another MO question we may assume that $X$ is affine and it is enough to prove that $H^i_x(X,\mathscr F)=0$ for $i=0,1$ for all $x\in X$ and it also follows that $\mathrm{depth}_Z\mathscr F\geq 2$ even if $Z$ is not contained in an affine piece of $X$. To do that write $\mathscr F^\vee$ as the quotient of a (locally) free sheaf ($X$ is affine!). Then $\mathscr F$ is a submodule of the dual of this locally free sheaf, let's call it $\mathscr E$, and the quotient $\mathscr E/\mathscr F$ is torsion-free. Therefore none of them have torsion and so $H^0_x(X,\mathscr F)=0$ and $H^1_x(X,\mathscr F)$ embeds into $H^1_x(X,\mathscr E)$. But the latter is $0$ by the assumption that $X$ is $S_2$.

EDIT 1 removed intro paragraph about the starting assumption.

EDIT 2 added "more generally" paragraph.

EDIT 3 added noetherian assumption. this is probably not necessary but without this one should possibly be more careful about the other conditions.

EDIT 4 Added Sketch above.

The assumption implies that the natural embedding induces an isomorphism $\Gamma(X,\mathscr O_X)\simeq \Gamma(\overline X,\mathscr O_{\overline X})$. Then this means that the complement of $X$ has at least codimension $2$.

In addition assume that $X$ is noetherian and $S_2$ (for instance normal).

In this case if $\Omega_{\overline X}$ is a reflexive sheaf, then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{\overline X})=\Gamma(X,\Omega_{X})$ is an isomorphism.

More generally, let $Z:=\overline X\setminus X$. If $\mathrm{depth}_Z\Omega_{\overline X}\geq 2$ then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{X})$ is an isomorphism. This certainly holds if $\Omega_{\overline X}$ is a reflexive sheaf, but obviously it could hold "by accident" even if one of the above conditions fail, so I am not claiming that these are necessary conditions, but at least they sure seem to provide a natural set of conditions under which the required map is an isomorphism.

Sketch that if $X$ is $S_2$, then a reflexive coherent sheaf $\mathscr F$ is also $S_2$: First observe that by the argument in this answer to another MO question we may assume that $X$ is affine and it is enough to prove that $H^i_x(X,\mathscr F)=0$ for $i=0,1$ for all $x\in X$. To do that write $\mathscr F^\vee$ as the quotient of a (locally) free sheaf ($X$ is affine!). Then $\mathscr F$ is a submodule of the dual of this locally free sheaf, let's call it $\mathscr E$, and the quotient $\mathscr E/\mathscr F$ is torsion-free. Therefore none of them have torsion and so $H^0_x(X,\mathscr F)=0$ and $H^1_x(X,\mathscr F)$ embeds into $H^1_x(X,\mathscr E)$. But the latter is $0$ by the assumption that $X$ is $S_2$.

EDIT 1 removed intro paragraph about the starting assumption.

EDIT 2 added "more generally" paragraph.

EDIT 3 added noetherian assumption. this is probably not necessary but without this one should possibly be more careful about the other conditions.

EDIT 4 Added Sketch above.

The assumption implies that the natural embedding induces an isomorphism $\Gamma(X,\mathscr O_X)\simeq \Gamma(\overline X,\mathscr O_{\overline X})$. Then this means that the complement of $X$ has at least codimension $2$.

In addition assume that $X$ is noetherian and $S_2$ (for instance normal).

In this case if $\Omega_{\overline X}$ is a reflexive sheaf, then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{\overline X})=\Gamma(X,\Omega_{X})$ is an isomorphism.

More generally, let $Z:=\overline X\setminus X$. If $\mathrm{depth}_Z\Omega_{\overline X}\geq 2$ then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{X})$ is an isomorphism. This certainly holds if $\Omega_{\overline X}$ is a reflexive sheaf, but obviously it could hold "by accident" even if one of the above conditions fail, so I am not claiming that these are necessary conditions, but at least they sure seem to provide a natural set of conditions under which the required map is an isomorphism.

Sketch that if $X$ is $S_2$, then a reflexive coherent sheaf $\mathscr F$ is also $S_2$: First observe that by the argument in this answer to another MO question we may assume that $X$ is affine and it is enough to prove that $H^i_x(X,\mathscr F)=0$ for $i=0,1$ for all $x\in X$ and it also follows that $\mathrm{depth}_Z\mathscr F\geq 2$ even if $Z$ is not contained in an affine piece of $X$. To do that write $\mathscr F^\vee$ as the quotient of a (locally) free sheaf ($X$ is affine!). Then $\mathscr F$ is a submodule of the dual of this locally free sheaf, let's call it $\mathscr E$, and the quotient $\mathscr E/\mathscr F$ is torsion-free. Therefore none of them have torsion and so $H^0_x(X,\mathscr F)=0$ and $H^1_x(X,\mathscr F)$ embeds into $H^1_x(X,\mathscr E)$. But the latter is $0$ by the assumption that $X$ is $S_2$.

EDIT 1 removed intro paragraph about the starting assumption.

EDIT 2 added "more generally" paragraph.

EDIT 3 added noetherian assumption. this is probably not necessary but without this one should possibly be more careful about the other conditions.

EDIT 4 Added Sketch above.

The assumption implies that the natural embedding induces an isomorphism $\Gamma(X,\mathscr O_X)\simeq \Gamma(\overline X,\mathscr O_{\overline X})$. Then this means that the complement of $X$ has at least codimension $2$.

In addition assume that $X$ is noetherian and $S_2$ (for instance normal).

In this case if $\Omega_{\overline X}$ is a reflexive sheaf, then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{\overline X})=\Gamma(X,\Omega_{X})$ is an isomorphism.

More generally, let $Z:=\overline X\setminus X$. If $\mathrm{depth}_Z\Omega_{\overline X}\geq 2$ then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{X})$ is an isomorphism. This certainly holds if $\Omega_{\overline X}$ is a reflexive sheaf, but obviously it could hold "by accident" even if one of the above conditions fail, so I am not claiming that these are necessary conditions, but at least they sure seem to provide a natural set of conditions under which the required map is an isomorphism.

EDIT 1 removed intro paragraph about the starting assumption.

EDIT 2 added "more generally" paragraph.

EDIT 3 added noetherian assumption. this is probably not necessary but without this one would possibly be more careful about the other conditions.

The assumption implies that the natural embedding induces an isomorphism $\Gamma(X,\mathscr O_X)\simeq \Gamma(\overline X,\mathscr O_{\overline X})$. Then this means that the complement of $X$ has at least codimension $2$.

In addition assume that $X$ is noetherian and $S_2$ (for instance normal).

In this case if $\Omega_{\overline X}$ is a reflexive sheaf, then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{\overline X})=\Gamma(X,\Omega_{X})$ is an isomorphism.

More generally, let $Z:=\overline X\setminus X$. If $\mathrm{depth}_Z\Omega_{\overline X}\geq 2$ then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{X})$ is an isomorphism. This certainly holds if $\Omega_{\overline X}$ is a reflexive sheaf, but obviously it could hold "by accident" even if one of the above conditions fail, so I am not claiming that these are necessary conditions, but at least they sure seem to provide a natural set of conditions under which the required map is an isomorphism.

Sketch that if $X$ is $S_2$, then a reflexive coherent sheaf $\mathscr F$ is also $S_2$: First observe that by the argument in this answer to another MO question we may assume that $X$ is affine and it is enough to prove that $H^i_x(X,\mathscr F)=0$ for $i=0,1$ for all $x\in X$. To do that write $\mathscr F^\vee$ as the quotient of a (locally) free sheaf ($X$ is affine!). Then $\mathscr F$ is a submodule of the dual of this locally free sheaf, let's call it $\mathscr E$, and the quotient $\mathscr E/\mathscr F$ is torsion-free. Therefore none of them have torsion and so $H^0_x(X,\mathscr F)=0$ and $H^1_x(X,\mathscr F)$ embeds into $H^1_x(X,\mathscr E)$. But the latter is $0$ by the assumption that $X$ is $S_2$.

EDIT 1 removed intro paragraph about the starting assumption.

EDIT 2 added "more generally" paragraph.

EDIT 3 added noetherian assumption. this is probably not necessary but without this one should possibly be more careful about the other conditions.

EDIT 4 Added Sketch above.