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Do the following Riemannian metrics on $GL(n,\mathbb{R})$ give us isometric structures?Do they generate the same volume forms? Is $O(n)$ a totally geodesic submanifold with respect to these metrics?

1. The metric with orthonormal frame $A\otimes A$ at each point $A\in GL(n,\mathbb{R})$

2. The metric with orthonormal frame $A\otimes A^{tr}$ for $A\in GL(n,\mathbb{R})$

Note that the tangent space $T_A GL(n,\mathbb{R})$ at each point $A$ is identified with $M_{n}(\mathbb{R})$ hence with $\mathbb{R}^{n^2}$.(Note that $GL(n,\mathbb{R})$ is counted as an open subset of Euclidean space hence has trivial tangent bundle so the above identification has an obvious desribtion). So at each point $A\in GL(n,\mathbb{R})$, each column of the tensor product matrix $A\otimes A$ (or $A\otimes A^{tr}$) can be considered as a tangent vector in $T_A GL(n,\mathbb{R})$. So $A\otimes A$ is obviously a fram, that is a base for the tangent space.This frame define a unique well defined Riemannian metric on $GL(n,\mathbb{R})$.

Edit: According to the comments on this question we clarify the identification $M_n(\mathbb{R})$ 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 $$(a_{11}, a_{12},a_{21}, a_{22})$$

Do the following Riemannian metrics on $GL(n,\mathbb{R})$ give us isometric structures?Do they generate the same volume forms? Is $O(n)$ a totally geodesic submanifold with respect to these metrics?

1. The metric with orthonormal frame $A\otimes A$ at each point $A\in GL(n,\mathbb{R})$

2. The metric with orthonormal frame $A\otimes A^{tr}$ for $A\in GL(n,\mathbb{R})$

Note that the tangent space $T_A GL(n,\mathbb{R})$ at each point $A$ is identified with $M_{n}(\mathbb{R})$ hence with $\mathbb{R}^{n^2}$.(Note that $GL(n,\mathbb{R})$ is counted as an open subset of Euclidean space hence has trivial tangent bundle so the above identification has an obvious desribtion). So at each point $A\in GL(n,\mathbb{R})$, each column of the tensor product matrix $A\otimes A$ (or $A\otimes A^{tr}$) can be considered as a tangent vector in $T_A GL(n,\mathbb{R})$. So $A\otimes A$ is obviously a fram, that is a base for the tangent space.This frame define a unique well defined Riemannian metric on $GL(n,\mathbb{R})$.

Edit: According to the comments on this question we clarify the identification $M_n(\mathbb{R})$ 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 $$(a_{11}, a_{12},a_{21}, a_{22})$$

Remark: For a homotopic version of this question please see this MO post.

Do the following Riemannian metrics on $GL(n,\mathbb{R})$ give us isometric structures?Do they generate the same volume forms? Is $O(n)$ a totally geodesic submanifold with respect to these metrics?

1. The metric with orthonormal frame $A\otimes A$ at each point $A\in GL(n,\mathbb{R})$

2. The metric with orthonormal frame $A\otimes A^{tr}$ for $A\in GL(n,\mathbb{R})$

Note that the tangent space $T_A GL(n,\mathbb{R})$ at each point $A$ is identified with $M_{n^2}(\mathbb{R})$ hence with $\mathbb{R}^{n^2}$. So at each point $A\in GL(n,\mathbb{R})$, each column of the tensor product matrix $A\otimes A$ (or $A\otimes A^{tr}$) can be considered as a tangent vector in $T_A GL(n,\mathbb{R})$. So $A\otimes A$ is obviously a fram, that is a base for the tangent space.This frame define a unique well defined Riemannian metric on $GL(n,\mathbb{R})$.

Edit: According to the comments on this question we clarify the identification $M_n(\mathbb{R})$ 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 $$(a_{11}, a_{12},a_{21}, a_{22})$$

Do the following Riemannian metrics on $GL(n,\mathbb{R})$ give us isometric structures?Do they generate the same volume forms? Is $O(n)$ a totally geodesic submanifold with respect to these metrics?

1. The metric with orthonormal frame $A\otimes A$ at each point $A\in GL(n,\mathbb{R})$

2. The metric with orthonormal frame $A\otimes A^{tr}$ for $A\in GL(n,\mathbb{R})$

Note that the tangent space $T_A GL(n,\mathbb{R})$ at each point $A$ is identified with $M_{n}(\mathbb{R})$ hence with $\mathbb{R}^{n^2}$.(Note that $GL(n,\mathbb{R})$ is counted as an open subset of Euclidean space hence has trivial tangent bundle so the above identification has an obvious desribtion). So at each point $A\in GL(n,\mathbb{R})$, each column of the tensor product matrix $A\otimes A$ (or $A\otimes A^{tr}$) can be considered as a tangent vector in $T_A GL(n,\mathbb{R})$. So $A\otimes A$ is obviously a fram, that is a base for the tangent space.This frame define a unique well defined Riemannian metric on $GL(n,\mathbb{R})$.

Edit: According to the comments on this question we clarify the identification $M_n(\mathbb{R})$ 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 $$(a_{11}, a_{12},a_{21}, a_{22})$$

Do the following Riemannian metrics on $GL(n,\mathbb{R})$ give us isometric structures?Do they generate the same volume forms? Is $O(n)$ a totally geodesic submanifold with respect to these metrics?

1. The metric with orthonormal frame $A\otimes A$ at each point $A\in GL(n,\mathbb{R})$

2. The metric with orthonormal frame $A\otimes A^{tr}$ for $A\in GL(n,\mathbb{R})$

Note that the tangent space $T_A GL(n,\mathbb{R})$ at each point $A$ is identified with $M_{n^2}(\mathbb{R})$ hence with $\mathbb{R}^{n^2}$. So at each point $A\in GL(n,\mathbb{R})$, each column of the tensor product matrix $A\otimes A$ (or $A\otimes A^{tr}$) can be considered as a tangent vector in $T_A GL(n,\mathbb{R})$. So $A\otimes A$ is obviously a fram, that is a base for the tangent space.This frame define a unique well defined Riemannian metric on $GL(n,\mathbb{R})$.

Edit: According to the comments on this question we clarify the identification $M_n(\mathbb{R})$ 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 $$(a_{11}, a_{12},a_{21}, a_{22})$$

A homotopy question about the above two frames: Are the two frames discussed in this questions, two homotopic frames(As two sections of the frame bundle of $GL(n,\mathbb{R})$?

Do the following Riemannian metrics on $GL(n,\mathbb{R})$ give us isometric structures?Do they generate the same volume forms? Is $O(n)$ a totally geodesic submanifold with respect to these metrics?

1. The metric with orthonormal frame $A\otimes A$ at each point $A\in GL(n,\mathbb{R})$

2. The metric with orthonormal frame $A\otimes A^{tr}$ for $A\in GL(n,\mathbb{R})$

Note that the tangent space $T_A GL(n,\mathbb{R})$ at each point $A$ is identified with $M_{n^2}(\mathbb{R})$ hence with $\mathbb{R}^{n^2}$. So at each point $A\in GL(n,\mathbb{R})$, each column of the tensor product matrix $A\otimes A$ (or $A\otimes A^{tr}$) can be considered as a tangent vector in $T_A GL(n,\mathbb{R})$. So $A\otimes A$ is obviously a fram, that is a base for the tangent space.This frame define a unique well defined Riemannian metric on $GL(n,\mathbb{R})$.

Edit: According to the comments on this question we clarify the identification $M_n(\mathbb{R})$ 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 $$(a_{11}, a_{12},a_{21}, a_{22})$$