Is there a vector field $X$ on $\mathrm{M}_n(\mathbb{R})$ or $\mathrm{GL}(n,\mathbb{R})$ with the following condition: $$\begin{cases} X\cdot \mathrm{trace}=\mathrm{Det} \\X\cdot \mathrm{Det}=-trace \end{cases}$$ where $\mathrm{Det}$ is determinant?

Is there a vector field $X$ on $\operatorname{M}_n(\mathbb{R})$ or $\operatorname{GL}(n,\mathbb{R})$ with the following condition: $$\begin{cases} X\cdot \operatorname{trace}=\operatorname{Det} \\X\cdot \operatorname{Det}=-\operatorname{trace} \end{cases}$$ where $\operatorname{Det}$ is determinant?

Is there a vector field $X$ on $M_n(\mathbb{R})$ or $GL(n,\mathbb{R})$ with the following condition: $$\begin{cases} X\cdot trace=Det \\X\cdot Det=-trace \end{cases}$$ where $Det$ is determinant?

Is there a vector field $X$ on $\mathrm{M}_n(\mathbb{R})$ or $\mathrm{GL}(n,\mathbb{R})$ with the following condition: $$\begin{cases} X\cdot \mathrm{trace}=\mathrm{Det} \\X\cdot \mathrm{Det}=-trace \end{cases}$$ where $\mathrm{Det}$ is determinant?

Is there a vector field $X$ on $M_n(\mathbb{R})$ or $GL(n,\mathbb{R})$ with the following condition: $$\begin{cases} X.trace=Det \\X.Det=-trace \end{cases}$$ where $Det$ is determinant?

Is there a vector field $X$ on $M_n(\mathbb{R})$ or $GL(n,\mathbb{R})$ with the following condition: $$\begin{cases} X\cdot trace=Det \\X\cdot Det=-trace \end{cases}$$ where $Det$ is determinant?