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Let $R$ be a ring, and $v\in R^2$ a vector. Let $G := \text{Stab}_{\text{GL}_2(R)}(v)$. Is $G^T$ (the group of transposes of matrices in $G$) the stabilizer of something? (Here, in both cases $GL_2(R)$ is acting on 'column vectors' by "left multiplication")

Ie, is there a vector $w\in R^2$ such that $G^T = \text{Stab}_{\text{GL}_2(R)}(w)$?

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    The matrices in $G^{T}$ will stabilize the row vector $v^{T}$ under right multiplication. – 2017-01-30
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    (In fact this is the best way to think of the transpose operation, it is mapping your left group action to a right group action.) – 2017-01-30
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    I don't understand your comments, @MorganRodgers . The OP has clearly indicated that $GL_2(\mathbb R)$ is acting on the vectors by left multiplication. – 2017-01-30
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    @user1551 I know. But changing the matrices for their transpose does not have a meaningful interpretation in this context. So I'm describing the context where this operation has meaning (as a comment, not an answer, since it doesn't answer the question; the answer to the question is "no"). – 2017-01-30
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    @MorganRodgers You say it doesn't have a meaningful interpretation in this context, but I don't see any context in the question. I think it can be given an interpretation: it's the dual rep of $G$ (essentially). Even without the interpretation I would have considered the question algebraically interesting on its own (at least if it didn't have a trivial answer). – 2017-01-30
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    @arctictern I would have loved to see that in your answer. – 2017-01-30

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No. For instance consider $R=\mathbb{R}$ and $v=[\begin{smallmatrix}1\\0\end{smallmatrix}]$.