matrix multiplied by rotation matrix on right side and transpose(rotation) on left side Would a matrix remain un-rotated if it is multiplied by an orthonormal rotation matrix on right side and transpose of same rotation matrix on the left side?
 A: As @hardmath suggested, by doing what you say you'd get a similarity transformation: suppose that you have a matrix $A$ and an invertible matrix $P$. Then, the matrix $B = P^{-1}AP$ is said to be similar to A.
And here's a counterexample. Suppose that you have the Rotation matrix
$R = \begin{bmatrix}
       \frac{\sqrt{3}}{2} & \frac{1}{2}          \\[0.3em]
       -\frac{1}{2} & \frac{\sqrt{3}}{2}            \\[0.3em] 
     \end{bmatrix}$
and its inverse (transpose)
$R^{-1} = \begin{bmatrix}
       \frac{\sqrt{3}}{2} & -\frac{1}{2}          \\[0.3em]
       \frac{1}{2} & \frac{\sqrt{3}}{2}            \\[0.3em] 
     \end{bmatrix}$
Then, if you consider a matrix 
$A = \begin{bmatrix}
       1 & 1          \\[0.3em]
       1 & 0            \\[0.3em] 
     \end{bmatrix}$
and you left-multiply it by $R^{-1}$ and right-multiply it by $R$, you get 
$\begin{bmatrix}
       \frac{\sqrt{3}}{4}(-2+\sqrt{3}) & \frac{1}{4}(2+\sqrt{3})          \\[0.3em]
       \frac{1}{4}(-1+\sqrt{3}(1+\sqrt{3})) & \frac{1}{4}(1+ 2\sqrt{3})            \\[0.3em] 
     \end{bmatrix}$
which is clearly a completely different matrix!
