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$\newcommand{\nN}{{\mathcal N}} \newcommand{\SL}{{\rm SL}} \newcommand{\G}{{\bf G}} \newcommand{\Q}{{\Bbb Q}} \newcommand{\R}{{\Bbb R}} \newcommand{\C}{{\Bbb C}} $The answer is No.

Write $\nN$ for the normalizer of $\G$ in $\SL_{8,\Q}$. We wish to find an anisotropic $\Q$-subgroup $\G'\in\SL_{8,\Q}$ such that $\G'_\R$ is conjugate to $\G_\R$ under $\SL(8,\R)$. Then $\G'\simeq{} _c \G$ for some 1-cocycle $c\in Z^1(\Q,\nN)$ such that $c$ becomes a coboundary in $ Z^1(\R,\nN)$. Since $\nN$ acts on $\G$ by inner automorphisms (see a comment of @YCor above), we see that $\G'$ must be an anisotropic inner form of $\G$ with an 8-dimensional representation defined over $\Q$.

Any inner form $\G'$ of $\G=\SL_{4,\Q}$ is isomorphic to $\SL(1,D)$ where $D$ is a central simple algebra of degree 4 over $\Q$. Since $\G'$ is anisotropic, $D$ is a division algebra. However, a representation $\rho$ of $\G'=\SL(1,D)$ over $\Q$ such that $\rho\otimes_\Q\C$ contains the standard 4-dimensional representation of $\G'_\C\simeq\SL_{4,\C}$ in $\C^4$, has dimension at least 16. Thus there is no anisotropic $\Q$-subgroup $\G'\subset\SL_{8,\Q}$ conjugate to $\G$ over $\R$ or over $\C$.

$\newcommand{\nN}{{\mathcal N}} \newcommand{\SL}{{\rm SL}} \newcommand{\G}{{\bf G}} \newcommand{\Q}{{\Bbb Q}} \newcommand{\R}{{\Bbb R}} \newcommand{\C}{{\Bbb C}} $The answer is No.

Write $\nN$ for the normalizer of $\G$ in $\SL_{8,\Q}$. We wish to find an anisotropic $\Q$-subgroup $\G'\in\SL_{8,\Q}$ such that $\G'_\R$ is conjugate to $\G_\R$ under $\SL(8,\R)$. Then $\G'\simeq{} _c \G$ for some 1-cocycle $c\in Z^1(\Q,\nN)$ such that $c$ becomes a coboundary in $ Z^1(\R,\nN)$. Since $\nN$ acts on $\G$ by inner automorphisms (see a comment of @YCor above), we see that $\G'$ must be an anisotropic inner form of $\G$ with an 8-dimensional representation defined over $\Q$.

Any inner form $\G'$ of $\G=\SL_{4,\Q}$ is isomorphic to $\SL(1,D)$ where $D$ is a central simple algebra of degree 4 over $\Q$. Since $\G'$ is anisotropic, $D$ is a division algebra. However, a representation $\rho$ of $\G'=\SL(1,D)$ over $\Q$ such that $\rho\otimes_\Q\C$ contains the standard 4-dimensional representation of $\G'_\C\simeq\SL_{4,\C}$ in $\C^4$, has dimension at least 16. Thus there is no anisotropic $\Q$-subgroup $\G'\subset\SL_{8,\Q}$ conjugate to $\G$ over $\R$ or over $\C$.

$\newcommand{\nN}{{\mathcal N}} \newcommand{\SL}{{\rm SL}} \newcommand{\G}{{\bf G}} \newcommand{\Q}{{\Bbb Q}} \newcommand{\R}{{\Bbb R}} \newcommand{\C}{{\Bbb C}} $The answer is No.

Write $\nN$ for the normalizer of $\G$ in $\SL_{8,\Q}$. We wish to find an anisotropic $\Q$-subgroup $\G'\in\SL_{8,\Q}$ such that $\G'_\R$ is conjugate to $\G_\R$ under $\SL(8,\R)$. Then $\G'\simeq{} _c \G$ for some 1-cocycle $c\in Z^1(\Q,\nN)$ such that $c$ becomes a coboundary in $ Z^1(\R,\nN)$. Since $\nN$ acts on $\G$ by inner automorphisms (see a comment of @YCor above), we see that $\G'$ must be an anisotropic inner form of $\G$ with an 8-dimensional representation defined over $\Q$.

Any inner form $\G'$ of $\G=\SL_{4,\Q}$ is isomorphic to $\SL(1,D)$ where $D$ is a central simple algebra of degree 4 over $\Q$. Since $\G'$ is anisotropic, $D$ is a division algebra. However, then $\G'=\SL(1,D)$ has no 8-dimensional representation over $\Q$ containing over $\C$ the standard 4-dimensional representation of $\G'_\C=\SL_{4,\C}$. Thus there is no anisotropic $\Q$-subgroup $\G'\subset\SL_{8,\Q}$ conjugate to $\G$ over $\R$ or over $\C$.

$\newcommand{\nN}{{\mathcal N}} \newcommand{\SL}{{\rm SL}} \newcommand{\G}{{\bf G}} \newcommand{\Q}{{\Bbb Q}} \newcommand{\R}{{\Bbb R}} \newcommand{\C}{{\Bbb C}} $The answer is No.

Write $\nN$ for the normalizer of $\G$ in $\SL_{8,\Q}$. We wish to find an anisotropic $\Q$-subgroup $\G'\in\SL_{8,\Q}$ such that $\G'_\R$ is conjugate to $\G_\R$ under $\SL(8,\R)$. Then $\G'\simeq{} _c \G$ for some 1-cocycle $c\in Z^1(\Q,\nN)$ such that $c$ becomes a coboundary in $ Z^1(\R,\nN)$. Since $\nN$ acts on $\G$ by inner automorphisms (see a comment of @YCor above), we see that $\G'$ must be an anisotropic inner form of $\G$ with an 8-dimensional representation defined over $\Q$.

Any inner form $\G'$ of $\G=\SL_{4,\Q}$ is isomorphic to $\SL(1,D)$ where $D$ is a central simple algebra of degree 4 over $\Q$. Since $\G'$ is anisotropic, $D$ is a division algebra. However, a representation $\rho$ of $\G'=\SL(1,D)$ over $\Q$ such that $\rho\otimes_\Q\C$ contains the standard 4-dimensional representation of $\G'_\C\simeq\SL_{4,\C}$ in $\C^4$, has dimension at least 16. Thus there is no anisotropic $\Q$-subgroup $\G'\subset\SL_{8,\Q}$ conjugate to $\G$ over $\R$ or over $\C$.

$\newcommand{\nN}{{\mathcal N}} \newcommand{\SL}{{\rm SL}} \newcommand{\G}{{\bf G}} \newcommand{\Q}{{\Bbb Q}} \newcommand{\R}{{\Bbb R}} \newcommand{\C}{{\Bbb C}} $The answer is No.

Write $\nN$ for the normalizer of $\G$ in $\SL_{8,\Q}$. We wish to find an anisotropic $\Q$-subgroup $\G'\in\SL_{8,\Q}$ such that $\G'_\R$ is conjugate to $\G_\R$ under $\SL(8,\R)$. Then $\G'\simeq{} _c \G$ for some 1-cocycle $c\in Z^1(\Q,\nN)$ such that $c$ becomes a coboundary in $ Z^1(\R,\nN)$. Since $\nN$ acts on $\G$ by inner automorphisms (see a comment of @YCor above), we see that $\G'$ must be an anisotropic inner form of $\G$ with an 8-dimensional representation defined over $\Q$.

Any inner form $\G'$ of $\G=\SL_{4,\Q}$ is isomorphic to $\SL(1,D)$ where $D$ is a central simple algebra of degree 4 over $\Q$. Since $\G'$ is anisotropic, $D$ is a division algebra. However, then $\G'=\SL(1,D)$ has no 8-dimensional representation over $\Q$ containing over $\C$ the standard 4-dimensional representation of $\G'_\C=\SL_{4,\C}$. Thus there is no anisotropic $\Q$-subgroup $\G'\in\SL_{8,\Q}$ conjugate to $\G$ over $\R$ or over $\C$.

$\newcommand{\nN}{{\mathcal N}} \newcommand{\SL}{{\rm SL}} \newcommand{\G}{{\bf G}} \newcommand{\Q}{{\Bbb Q}} \newcommand{\R}{{\Bbb R}} \newcommand{\C}{{\Bbb C}} $The answer is No.

Write $\nN$ for the normalizer of $\G$ in $\SL_{8,\Q}$. We wish to find an anisotropic $\Q$-subgroup $\G'\in\SL_{8,\Q}$ such that $\G'_\R$ is conjugate to $\G_\R$ under $\SL(8,\R)$. Then $\G'\simeq{} _c \G$ for some 1-cocycle $c\in Z^1(\Q,\nN)$ such that $c$ becomes a coboundary in $ Z^1(\R,\nN)$. Since $\nN$ acts on $\G$ by inner automorphisms (see a comment of @YCor above), we see that $\G'$ must be an anisotropic inner form of $\G$ with an 8-dimensional representation defined over $\Q$.

Any inner form $\G'$ of $\G=\SL_{4,\Q}$ is isomorphic to $\SL(1,D)$ where $D$ is a central simple algebra of degree 4 over $\Q$. Since $\G'$ is anisotropic, $D$ is a division algebra. However, then $\G'=\SL(1,D)$ has no 8-dimensional representation over $\Q$ containing over $\C$ the standard 4-dimensional representation of $\G'_\C=\SL_{4,\C}$. Thus there is no anisotropic $\Q$-subgroup $\G'\subset\SL_{8,\Q}$ conjugate to $\G$ over $\R$ or over $\C$.