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Let $G$ be a subgroup of $\mathrm{GL}(n,\mathbb{F})$. Denote by $G^T$ the set of transposes of all elements in $G$. Can we always find an $M\in \mathrm{GL}(n,\mathbb{F})$ such that $A\mapsto M^{-1}AM$ is a well-defined map from $G$ to $G^T$?

For example if $G=G^T$ then any $M\in G$ will do the job.

Another example, let $U$ be the set of all invertible upper triangular matrices. Take $M=(e_n\,\cdots\,e_2\,e_1)$ where $e_i$ are the column vectors that make $I=(e_1\,e_2\,\cdots\,e_n)$ into an identity matrix. Then $M$ do the job. For $n=3$, here what the $M$ will do $$\begin{pmatrix}a&b&c\\ 0&d&e\\ 0&0&f\end{pmatrix}\mapsto M^{-1}\begin{pmatrix}a&b&c\\ 0&d&e\\ 0&0&f\end{pmatrix} M=\begin{pmatrix}f&e&c\\0&d&b\\0&0&a\end{pmatrix}^T$$

Thank you.

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The answer is no, in general. For example, when ${\mathbb F}$ is the field of two elements, let $G$ be the stabilizer of the one-dimensional subspace of ${\mathbb F}^3,$ (viewed as column vectors, with ${\rm GL}(3,\mathbb{F})$ acting by left multiplication) consisting of vectors with $0$ in positions $2$ and $3$. Then $G \cong S_{4},$ but $G$ does not stabilize any $2$-dimensional subspace. However $G^{T}$ is the stabilizer of the $2$-dimensional subspace consisting of vectors with $0$ in position $1$. Hence the subgroups $G$ and $G^{T}$ are not conjugate within ${\rm GL}(3,\mathbb{F}).$

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Thank you for your answer. I am not an expert in this, so I need some clarification. Could you tell me which one-dimensional subspace of $\mathbb{F}^3$ stabilized by $G$ (or doesn't matter which one-dimensional subspace that we choose?). Why $G$ does not stabilize any $2$-dimensional subspace? –  user9077 Mar 2 '12 at 20:27
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