After thinking about Nick P's answer and re-reading the relevant chapter of Sussman's [*Structure and Interpretation of Classical Mechanics*][1], 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][3].

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)][2] is,

$$D_s H(t,q,p) = (1,\partial_2 H(t,q,p),-\partial_1 H(t,q,p)).$$

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),

$$D_s H \circ C = DC \cdot D_s H'.$$

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 $H' = H \circ C$.


  [1]: http://mitpress.mit.edu/sites/default/files/titles/content/sicm/book.html
  [2]: http://mitpress.mit.edu/sites/default/files/titles/content/sicm/book-Z-H-59.html#%25_sec_5.2
  [3]: http://mitpress.mit.edu/sites/default/files/titles/content/sicm/book-Z-H-79.html#%_chap_8