How to understand the geometry of $A \in O(n,\mathbb{C})$ versus $A \in U(n)$ How to understand the geometry of $A \in O(n,\mathbb{C})$ versus $A \in U(n)$

So if $A \in O(n,\mathbb{R})$ then $A$ will act as a combination of rotations and reflections. If $A \in SO(n,\mathbb{R})$ then $A$ will act just as a rotation.
Is there a straightforward interpretation for:
$$A \in O(n,\mathbb{C})$$
$$A \in SO(n,\mathbb{C})$$
$$A \in U(n)$$
I suppose that $A \in O(n,\mathbb{C})$ and $A \in SO(n,\mathbb{C})$ would be analogous to the real case, it's just now the matrices are reflecting and rotating the complex plane (edit: NOT TRUE see Runway44's answer).
But then what about $A \in U(n)$?
I know that $A \in U(n) \rightarrow det(A) = e^{i \theta}$. I understand the determinant of the linear transformation as giving the scaling factor for the volume, but I don't know how to interpret this if the determinant is a complex value.
Kind of some open ended questions here, any insights would be appreciated. Thanks.

edit:

*

*Isoclinic rotations are rotations $\varphi$ in $\mathbb{R}^{2n}$ such that there exists $n$ complementary oriented planes $P_i=\langle x_i,y_i\rangle$ such that $\varphi$ acts as a simple rotation of the same angle $\theta$ on each plane $P_i$, either clockwise or counter-clockwise (here we assume that an orientation of $\mathbb{R}^{2n}$ has been chosen, and that the orientations of the $P_i$ are chosen such that it induces the same orientation on $\mathbb{R}^{2n}$). Are left isoclinic rotations a group?
 A: I think you dismissed the harder case too early and got hung up on the easier case.
By the spectral theorem, every unitary matrix is diagonalizable with eigenvalues of the form $e^{i\theta}$. So a unitary matrix acts as a phasor multiplication in a bunch of perpendicular complex lines. If we write $\mathbb{C}^n\cong\mathbb{R}^{2n}$, multiplication-by-$i$ is a particular isoclinic rotation of $\mathbb{R}^{2n}$.  Our "complex lines" are 2D subspaces which are stable under this distinguished right-angle isoclinic rotation, and two "complex lines" are perpendicular if after applying the isoclinic rotation with any two angles to the two planes they are still orthogonal (with respect to the real inner product, which is the real part of the complex inner product). The planes stabilized by the isoclinic rotation ("complex lines") are parametrized by the complex projective space $\mathbb{CP}^{n-1}$.
For complex orthogonal groups, let's consider $SO(2,\mathbb{C})$. Its lie algebra $\mathfrak{so}(2,\mathbb{C})$ may be written as a direct sum of two vector subspaces, $\mathfrak{so}(2,\mathbb{R})\oplus i\mathfrak{so}(2,\mathbb{R})$. I expect, then, every element is expressible as a product $\exp(\theta J)\exp(i\tau J)$, where $J=[\begin{smallmatrix}0&-1\\1&\phantom{-}0\end{smallmatrix}]$. The former is just the usual rotation matrix, the latter is
$$ \exp(i\tau J)=\cos(i\tau)+\sin(i\tau)J=\cosh\tau+i\sinh\tau J=
\begin{bmatrix}\cosh\tau & -i\sinh\tau \\ i\sinh\tau & \cosh\tau\end{bmatrix} $$
The eigenvectors are $[\begin{smallmatrix}~1\\ \pm i\end{smallmatrix}]$ with eigenvalues $e^{\pm\tau}$. In other words, we can decompose $\mathbb{C}^2=\mathbb{R}^2\oplus i\mathbb{R}^2$; the matrix $\exp(i\theta)$ acts by the same rotation in both the real and imaginary parts. If we define the diagonal and antidiagonal parts $\{(v,v)\mid v\in\mathbb{R}^2\}$ and $\{(v,-v)\mid v\in\mathbb{R}^2\}$, the matrix $\exp(i\tau J)$ is a "squeeze map" which expands the diagonal part by $e^\tau$ and shrinks the anti-diagonal part by $e^{-\tau}$ (or vice-versa, depending on $\tau$'s sign).
Generalizing to $\mathrm{SO}(n,\mathbb{C})$, hopefully elements are expressible as $\exp(X)\exp(iY)$ with $X,Y\in\mathfrak{so}(n,\Bbb R)$. By the spectral theorem, $X$ and $Y$ are each a combination of a right-angle rotation and a real scaling in a bunch of orthogonal 2D planes (the stable planes for $X$ independent of the ones for $Y$), and what happened in the 2D situation happens for all of these planes.
I think the Cartan-Dieudonne theorem guarantees $\mathrm{O}(n,\mathbb{C})$ is generated by "pseudo" Householder matrices $I-2vv^T$ (for real-coordinate vectors $v$, this real matrix acting on $\mathbb{R}^n$ represents the reflection across the hyperplane perpendicular to $v$). So it'd be interesting if we could find a sensible geometric interpretation of $I-2vv^T$ as a transformation of $\mathbb{C}^n$.
