Assume we have a homomorphism $\phi: C(S^{1},M_{n}(\mathbb{C}))\rightarrow C(S^{1},M_{m}(\mathbb{C}))$ where $n$ divides $m$. Under what conditions does $\phi$ send constant functions to constant functions?

This is perhaps just a partial answer; but I feel that Will's answer is incomplete, and this is a more interesting question that he is suggesting. Given a *homomorphism $\phi$, composing with pointevaluation at any point $x \in S^1$ gives us a representation $$\phi_x: C(S^1, M_n) \to M_m, $$ which is therefore of the form $$ \phi_x(f) = \alpha_x(\text{diag}(f(y(x,1)), \dots, f(y(x,k)))), $$ where $\alpha_x$ is an automorphism of $M_m$ and $y(x,1),\dots,y(x,k) \in S^1$ (and $k=m/n$). Now, $\phi$ sends constant functions to constant functions if and only if $\alpha_x \circ \alpha_y^{1}$ acts as the identity on the $M_m \otimes 1_k \subset M_n$. I'm not sure if there is a more explicit characterization. Some remarks:



Basically never. Take a typical homomorphism, then conjugate it by a very nonconstant function in $C(S^1, GL_m(\mathbb C))$. Conjugation is always a homomorphism, and satisfies every niceness condition I can think of. Any constant matrix that become a constant nonscalar matrix after the first homomorphism will become a nonconstant matrix after conjugation. Even constant scalar matrices can become constant nonscalar matrices and then nonconstant matrices. For instance, you can send $M$ to $\left(\begin{array}{cc} M & 0 \\ 0 & 0 \end{array}\right)$ or $\left(\begin{array}{cc} M & 0 \\ 0 & \overline{M} \end{array}\right)$ But this can be avoided by requiring the homomorphism to be unital and complexlinear. 

