Let $M_n$ be the vector space of $n \times n$ real matrices.
We say a linear operator $\alpha:M_n \to M_n$ is hemitropic* if:
$(*) \, \, \alpha(S^TXS)=S^T\alpha(X)S \, , \, \forall S \in SO(n)$
and isotropic, if the above formula holds $\forall S \in O(n)$
My question: Is every hemitropic operator necessarily isotropic? Does the answer change if we assume $\alpha$ is injective?
In odd dimensions, the answer is yes, since for any $Q \in O(n) \setminus SO(n)$, $\det(-Q)=1$ so $-Q \in SO(n)$ and by (*):
I suspect the answer in even dimensions is negative, but so far I didn't find a way to construct a hemitropic non-isotropic operator. (Though I guess this can be done even in dimension $2$).
For dimension $2$ I have constructed (see answer below) a hemitropic non-isotropic operator. However, my construction relied on the fact that all $2$-dimensional rotation commute. This is not the case for higher dimensions. (See here).
This leaves open the question for even dimensions above two.
The terminology "hemitropic" comes from Elasticity theory.