Transformation that rotates eigenvalues Let $A \in \mathbb{R}^{n \times n}$ be a square matrix. 
Is there a transformation $T_{\theta}: \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^{n \times n}$, not necessarily linear, that rotates the eigenvalues by an angle $\theta$ on the complex plane?
In other words, for each eigenvalue of $A$, $\lambda(A) \in \mathbb{C}$, there is an eigenvalue of $T_{\theta}(A)$, $\lambda( T_{\theta}(A) ) \in \mathbb{C}$, such that $\lambda( T_{\theta}(A) ) = e^{j \theta} \lambda(A)$.

In general, you can't. The reason is that you want to keep real entries, hence a characteristic polynomial with real coefficients, and the complex roots of such a polynomial have to come in conjugate pairs. If you rotate the eigenvalues of $A$ arbitrarily, you lose this symmetry property and the set that you obtain is not the set of eigenvalues.
 of a real-entry matrix.

Edit: even without thinking of multiplicities, the set $S$ of eigenvalues of a real-entry matrix is symmetric with respect to the real axis. In other words a set $S'\subset \Bbb C$ that is not self-conjugate is not the set of eigenvalues of any real-entry matrix. 
If $A$ has at least one non-zero eigenvalue, then there are at most finitely many rotations $r_\theta : \Bbb C\to \Bbb C$ such that $r_\theta (S)$ still has this property. 
The only rotations that keep this property for all matrices are the identity and the rotation by $\pi$. They correspond to $T=\operatorname{Id}$ and $T=-\operatorname{Id}$ respectively. For instance, if $A=\operatorname{Id}$, the only eigenvalue is $1$; you can't hope to rotate it by $\theta\not\in\{0,\pi\}$.

If you allow for complex entries however, the obvious $B=e^{i\theta}A$ (multiply entry-wise) works.
