Homotopy classes of maps which commute with complex conjugation 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 $[M,N]_g$.
 A: As far as I can tell, what you denote by $[U(1),U(n)]_*$ does not have a natural group structure: let $\gamma$ and $\delta$ be continuous maps from the circle $U(1)$ to $U(n)$ commuting with complex conjugation, i.e., equivariant with respect to the action of the Galois group $G=\mathrm{Gal}(\mathbf C/\mathbf R)$. Moreover, assume that $\gamma$ and $\delta$ are base-point preserving. Then the usual concatenation $\gamma\delta$ of closed paths is not necessarily equivariant. Indeed, the path $\gamma\delta$ is equal to $\gamma$ on the upper hemicircle and equal to $\delta$ on the lower one and has no reason to be equivariant.
However, one can determine $[U(1),U(n)]_*$ as as set and it will turn out to be naturally bijective with the disjoint union $\mathbf Z\coprod\mathbf Z$ of two copies of the set of integers $\mathbf Z$, or if you prefer with the set $\mathbf Z/2\times\mathbf Z$.
A usual observation in equivariant homotopy theory is that an equivariant map $\gamma$ from the circle $U(1)$ to $U(n)$ is uniquely determined by its restriction to the closed upper hemicircle $U(1)_\geq$, and amounts to a continuous map from the interval $I=[-1,1]$ to $U(n)$ that maps the end points into the subset $O(n)$ of $U(n)$. When one considers base-point preserving maps $\gamma$, one has, moreover, $\gamma(1)=1$. A base-point preserving equivariant homotopy of base-point preserving equivariant maps from $U(1)$ to $U(n)$ corresponds to a homotopy of continuous maps from $I$ to $U(n)$ mapping $1$ to $1$, and $-1$ into $O(n)$ at all times. Briefly put, one has
$$
[U(1),U(n)]_*=[(I,\partial I),(U(n),O(n))].
$$
Assuming that $n\geq1$, one distinguishes two subsets of homotopy classes of such maps from $I$ into $U(n)$: the ones that map $-1$ to $SO(n)$, and the ones that map $-1$ to the other connected component $O^-(n)$ of $O(n)$. The former is nothing but
$$
[(I,\partial I),(U(n),SO(n))];
$$
we will denote the latter by
$$
[(I,-1),(U(n),O^-(n))].
$$
One can show that both subsets are in bijection with $\mathbf Z$.
As for the former set, the natural map
$$
\pi_1(U(n))\rightarrow[(I,\partial I),(U(n),SO(n))]
$$
is a bijection since $SO(n)$ is connected and the morphism $\pi_1(SO(n))\rightarrow\pi_1(U(n))$ is trivial. Here I consider $\pi_1(U(n))$ as the set of homotopy classes of pairs from $(I,\partial I)$ to $(U(n),1)$.
As for the latter set, choose a path $\gamma$ from $I$ to $U(n)$ such that $\gamma(1)=1$ and $\gamma(-1)\in O^-(n)$. The map
$$
\pi_1(U(n))\rightarrow[(I,-1),(U(n),O^-(n))]
$$
defined by concatenating with $\gamma$ is a bijection for similar reasons.
