Consider the group $U(2n)$ of unitary matrices. This has two standard and important homomorphic involutions. The most famous one $A \mapsto \overline A$ is complex conjugation, with fixed-set $O(2n)$, the set of real orthogonal matrices. But the one I'm most interested in here is conjugation-by-j: if one considers $\Bbb H^n = (\Bbb C \oplus j \Bbb C)^n = \Bbb R^{4n} $ then the matrix representing right-multiplication-by-j is the $4n \times 4n$ block-diagonal orthogonal matrix, given by summing n copies of $$\begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}.$$

Now conjugation by $j$ induces an outer automorphism of $U(2n)$ of order 2 whose fixed-set is $\text{Sp}(n)$, the compact group of quaternionic-linear unitary matrices.

It is well-known that unitary groups $U(2n)$ have perfect Morse functions (and I believe $\text{Sp}(n)$ does as well, but I didn't check). I'm interested in the $\Bbb Z/2$-equivariant Morse theory of $U(2n)$, and for computations it would be nice to have as small a complex as possible.

Question. What is the minimal number of $\Bbb Z/2$-critical orbits for a $j$-invariant Morse function on $U(2n)$? What techniques are available to prove that a number of critical orbits is optimal?

For my own work I am more immediately interested in the case $n = 1$, where a more concrete formulation can be given: $U(2)$ is equivariantly diffeomorphic to $S^1 \times S^3$ with involution $\iota(\lambda, z, w) = (\bar \lambda, \lambda z, \lambda w)$. One can cook up an invariant Morse functions with two critical points on $1 \times S^3$ and four critical orbits on $-1 \times S^3$, but I don't know if this is minimal.

I believe that the sum of the ranks of Bredon equivariant homology (with suitable coefficient systems: I think one needs to assume that $A(G/H)$ is rank 1 for all orbits) gives a lower bound for the number of critical orbits of an invariant Morse function. Based on my calculations, I don't think this is sufficient to show that my '6' is minimal in this case (I only seem to be able to prove a lower bound of 4), but it's very possible I made some calculation errors.