Geometric meaning of conjugation Given an $n\times n$ full rank real $M,P,R$ and orthogonal $Q$ what is the geometric meaning and relation/differences between following operations?
$$PMP^{-1}$$
$$PMR$$
$$QMQ'$$  
 A: *

*We say that two matrices $M,M' \in M_n(\mathbb{F})$ are similar if there exists an invertible matrix $P$ such that $M = P^{-1} M P$. Geometrically, similar matrices (can) represent the same linear transformation with respect to different bases. More precisely, $M$ and $M'$ are similar if and only if there exists an $n$-dimensional vector space $V$ over $\mathbb{F}$, a linear transformation $T \colon V \rightarrow V$ and two bases $\mathcal{B}_1, \mathcal{B}_2$ such that $M = [T]^{\mathcal{B}_1}_{\mathcal{B}_1}$ and $M' = [T]^{\mathcal{B}_2}_{\mathcal{B}_2}$.

*We say that two matrices $M, M' \in M_n(\mathbb{F})$ are row-and-column equivalent (or just eqivalent) if there exists invertible matrices $P,R$ such that $M = R M' P$. Algebraically, this means that $M'$ can be transformed to $M$ (and vice versa) by performing elementary row and columns operations or, more simply, that $M'$ and $M$ have the same rank. Geometrically, row-and-column equivalent matrices (can) represent the same linear transformation with respect to different bases both for the domain and range. More precisely, $M$ and $M'$ are row-and-column equivalent if and only if there exists an $n$-dimensional vector space $V$ over $\mathbb{F}$, a linear transformation $T \colon V \rightarrow V$ and four bases $\mathcal{B}_1,\mathcal{B}_1', \mathcal{B}_2, \mathcal{B}_2'$ such that $M = [T]^{\mathcal{B}_1}_{\mathcal{B}_1'}$ and $M' = [T]^{\mathcal{B}_2}_{\mathcal{B}_2'}$. Unlike the previous relation, this can be generalized to non-square matrices and linear maps $T \colon V \rightarrow W$ between two different vector spaces.

*We say that two matrices $M,M' \in M_n(\mathbb{R})$ are orthogonally similar if there exists an orthogonal matrix $O$ (necessarily invertible) such that $M = O^{-1} M O = O^T M O$. Geometrically, orthogonally similar matrices (can) represent the same linear transformation with respect to different orthonormal bases. More precisely, $M$ and $M'$ are orthogonally similar if and only if there exists an $n$-dimensional real inner product space $(V, \left< \cdot, \cdot \right>)$, a linear transformation $T \colon V \rightarrow V$ and two orthonormal bases $\mathcal{B}_1, \mathcal{B}_2$ such that $M = [T]^{\mathcal{B}_1}_{\mathcal{B}_1}$ and $M' = [T]^{\mathcal{B}_2}_{\mathcal{B}_2}$.

*Finally, We say that two matrices $M, M' \in M_n(\mathbb{R})$ are orthogonally equivalent if there exists orthogonal matrices $O,Q$ such that $M = O M' Q$.  Geometrically, orthogonally equivalent matrices (can) represent the same linear transformation with respect to different orthonormal bases both for the domain and range. More precisely, $M$ and $M'$ are orthogonally equivalent if and only if there exists a real $n$-dimensional inner product space $(V, \left< \cdot, \cdot \right>$, a linear transformation $T \colon V \rightarrow V$ and four orthonormal bases $\mathcal{B}_1,\mathcal{B}_1', \mathcal{B}_2, \mathcal{B}_2'$ such that $M = [T]^{\mathcal{B}_1}_{\mathcal{B}_1'}$ and $M' = [T]^{\mathcal{B}_2}_{\mathcal{B}_2'}$. Unlike the previous relation, this can be generalized to non-square matrices and linear maps $T \colon V \rightarrow W$ between two different real inner product spaces.

