After thinking about Nick P's answer and re-reading the relevant chapter of Sussman's Structure and Interpretation of Classical Mechanics, I came up with the following elaboration of Nick's argument. It's not water-tight, but it convinced me, and perhaps it will help someone else. I will use Sussman's unorthodox but precise notation.
The first step (and this is the part that I can't rigorously justify) is to expand the definition of the phase-space derivative operator. The definition given by Sussman in Eq. (5.15) is,
$$D_s H(t,p,q) = (1,\partial_2 H(t,q,p),-\partial_1 H(t,q,p)).$$$$D_s H(t,q,p) = (1,\partial_2 H(t,q,p),-\partial_1 H(t,q,p)).$$
The extension we'll make is to define $D_s$ for Hamiltonians that are functions of complex-conjugate coordinates and momenta as,
$$D_s H(t,\psi,\psi^*) = (1,-\imath \partial_2 H(t,\psi,\psi^*),\imath \partial_1 H(t,\psi,\psi^*)).$$
With this extension, Hamilton's equations can be written in the same form for both the usual real and the complex coordinates:
$$D \sigma = D_s H \circ \sigma,$$
where $\sigma(t) = (t, q(t), p(t))$ or $\sigma(t) = (t, \psi(t), \psi^*(t))$, a mapping from time to phase space position, represents a path.
Now, let $C$ be a phase space coordinate transformation: $\sigma = C \circ \sigma'$. The transformation is canonical if there exists a new Hamiltonian $H'$ such that the equations of motion derived from it describe the same motion of the system. A sufficient condition for this is Eq. (5.19),
$$D_s H \circ C = DC \circ D_s H'.$$$$D_s H \circ C = DC \cdot D_s H'.$$
I'll show that the transformation,
$$(t, \psi(t), \psi(t)^*) = C(t, n(t), \phi(t)) = (t, \sqrt{n} e^{\imath \phi}, \sqrt{n} e^{-\imath \phi})$$
satisfies this condition for any $H$, and that furthermore $H' = H \circ C$, i.e. the new Hamiltonian can be obtained from the old one simply by substituting $\sqrt{n} e^{\imath \phi}$ for $\psi$. (This doesn't generally have to be the case: for example, it's not the case for the transformation $\psi = \sqrt{n} e^{2\imath \phi}$ discussed by Nick.)
The left-hand-side of the sufficient condition is,
$$D_s H(t, \psi, \psi^*) = (1, -\imath \partial_2 H(t,\psi,\psi^*), \imath \partial_1 H(t,\psi,\psi^*))$$ $$(D_s H \circ C)(t, n, \phi) = (1, -\imath (\partial_2 H) \circ C, \imath (\partial_1 H) \circ C)$$
Here, $\partial_1$ is the partial derivative with respect to the first argument (i.e., $\psi$: following Sussman, I'm using zero-based indexing, where time is the zeroth argument).
On the right-hand-side, the Jacobian of the transformation is,
$$DC = [\partial_0 C, \partial_1 C, \partial_2 C] = \begin{bmatrix} \begin{pmatrix} \partial_{0,0} C \\ \partial_{0,1} C \\ \partial_{0,2} C \end{pmatrix} & \begin{pmatrix} \partial_{1,0} C \\ \partial_{1,1} C \\ \partial_{1,2} C \end{pmatrix} & \begin{pmatrix} \partial_{2,0} C \\ \partial_{2,1} C \\ \partial_{2,2} C \end{pmatrix} \end{bmatrix}$$
and so the right-hand-side reads,
$$DC \cdot D_s H' = \begin{pmatrix} \partial_{0,0} C + \partial_{1,0} C \partial_2 H' - \partial_{2,0} C \partial_1 H' \\ \partial_{0,1} C + \partial_{1,1} C \partial_2 H' - \partial_{2,1} C \partial_1 H' \\ \partial_{0,2} C + \partial_{1,2} C \partial_2 H' - \partial_{2,2} C \partial_1 H' \end{pmatrix}.$$
The only nonzero elements of the Jacobian are,
$$\partial_{1,2} C = \frac{1}{2\sqrt{n}} e^{-\imath \phi},$$ $$\partial_{2,2} C= -\imath \sqrt{n} e^{-\imath \phi},$$ $$\partial_{1,1} C = \frac{1}{2\sqrt{n}} e^{\imath \phi},$$ $$\partial_{2,1} C = \imath \sqrt{n} e^{\imath \phi},$$ $$\partial_{0,0} C = 1.$$
The canonicity condition reduces to the system of equations,
$$-\imath (\partial_2 H) \circ C = \partial_{1,1} C \partial_2 H' - \partial_{2,1} C \partial_1 H'$$ $$\imath (\partial_1 H) \circ C = \partial_{1,2} C \partial_2 H' - \partial_{2,2} C \partial_1 H'$$
Solving for $\partial_1 H'$ gives,
$$\imath (\partial_{2,2} C \partial_{1,1} C - \partial_{2,1} C \partial_{1,2} C) \partial_1 H' = \partial_{1,2} C (\partial_2 H) \circ C + \partial_{1,1} C (\partial_1 H) \circ C.$$
The quantity in parentheses on the left is exactly $-\imath$, so using the chain rule,
$$\partial_1 H' = ((\partial_2 H) \circ C ) \partial_{1,2} C + ((\partial_1 H) \circ C) \partial_{1,1} C = \partial_1 (H\circ C).$$
A similar relation holds for $\partial_2 H'$. Therefore, the transformation can be made canonical using the "natural" choice of $H' = H \circ C$.