Taking an orthonormal basis as $ \vec i, \vec j ,\vec k$ the rotation tensor $T$ around a certain axis is represented by an orthogonal matrix $ M$.
If I choose any other kind of basis will the matrix that represents the same tensor $T$ be orthogonal?
Rotation in the plane by 90 degrees counterclockwise under canonical coordinates is given by the matrix $$\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$
Now consider the coordinate system with basis vectors $\vec i = \begin{bmatrix}1&0\end{bmatrix}^T$ and $\vec h = \begin{bmatrix}1&1\end{bmatrix}^T$. Under this rotation $\vec i \mapsto \begin{bmatrix}0&1\end{bmatrix}^T = \vec h - \vec i$ and $\vec h \mapsto \begin{bmatrix}-1&1\end{bmatrix}^T = \vec h - 2\vec i$.
So the rotation matrix for this coordinate system is $$\begin{bmatrix}-1&1\\-2&1\end{bmatrix}$$ which is clearly not orthogonal.
More generally, for any change of basis, there is some change-of-basis matrix $P$ that converts from canonical coordinates to the new coordinates. The only condition $P$ has to meet to be a change-of-basis matrix is to be invertible. The rotation matrix in the new coordinate system can be obtained by using $P^{-1}$ to convert from the new coordinates to canonical coordinates, applying $M$ to rotate in canonical coordinates, then applying $P$ to convert back to the new coordinate system: $$\tilde M = PMP^{-1}$$
The condition of being orthogonal is $M^TM = I$. But $$\tilde M^T\tilde M = {P^{-1}}^TM^TP^TPMP^{-1}$$ And if $P$ is not orthogonal, there is no reason to expect that to reduce to $I$.