If we consider all continuous maps $$f : U(1) \to U(n),$$we know the homotopy classes form a group $[U(1),U(n)] = \pi_1(U(n)) = \mathbb Z$.
However, suppose we only want to consider maps which commute with complex conjugation, i.e. $f(e^{-i\alpha}) = f(e^{i\alpha})^*$ (where $^*$ denotes complex conjugation, not Hermitian conjugation). What is then the group (or set) of homotopy classes?
One might denote this as $[U(1),U(n)]_*$, and I expect that $$[U(1),U(n)]_* = \mathbb Z_2 \times \mathbb Z$$ (where the $\mathbb Z_2$ factor corresponds to the fact that $f(1) \in O(n)$, and $\pi_0(O(n)) = \mathbb Z_2$) but I am not sure how to prove this.
I realize there is a more general question lurking in the background, which must have been treated extensively somewhere. The general formulation would be: consider manifolds $M,N$. By $[M,N]$ we would denote the set of all homotopy classes for continuous functions $f:M\to N$. Suppose there are fixed continuous functions $g_M: M \to M$ and $g_N: N \to N$. What is then the set of homotopy classes of continuous functions $f:M \to N$ which satisfy $f \circ g_M = g_N \circ f$? One might denote this as $[M,N]_g$.