Put $M=GL(n,\mathbb{R})$Edit: According to comment of Prof.Assume that $n^2-n$ is not divided by $4$ otherwise GoodWillie we revise the answer to this question is obviuosely "yes" since the mapping.
Put $A\mapsto A^{tr}$ is homotopic to identity$M=GL(n,\mathbb{R})$.
We identify $M_n(\mathbb{R})$ with $\mathbb{R}^{n^2}$: The identification is based on the lexicographic order on the index $i,j$ in $(a_{ij})$. For example $$ \begin{pmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}$$
is identified with $$(a_{11}, a_{12},a_{21}, a_{22})$$
So $M$ being an open subset of $\mathbb{R}^{n^2}$ has trivial tangent bundle and there is an obvious description for a tangent vector at a point $A\in M$.
The mapping $A\mapsto A\otimes A$ defines a section of fram bundle of $M$.Because each row of $A\otimes A$ is counted as a $n×n$ matrix via the above identification. So $A\otimes A$ actualy determines $n^2$ independent vectors in the tangent space to $M$ at $A$. In a similar manner $A\otimes A^{tr}$ is another section of the frame bundle of manifold $M$ where $tr$ is transpose operator.
Are the above two sections $A\otimes A$ and $A\otimes A^{tr}$ homotopic sections of frame bundle of $M$?