I'm not really sure about your notation, so I'm going to assume that $\mathbf{i}$ = $\vec{x}$ and $\mathbf{j}$ = $\vec{y}$ in the lab frame (I think I've seen this in physics textbooks before). Assuming that's the case, the transform from the lab frame to the rotating frame is:
$$\vec{x} = \vec{x}'\cos{\omega_0t} - \vec{y}'\sin{\omega_{0}t}$$
$$\vec{y} = \vec{x}'\sin{\omega_0t} + \vec{y}'\cos{\omega_{0}t}$$
So if you take your equation:
$$\mathbf{H_{1}} = H_1\left(\vec{x}\cos{\omega_{z}t} + \vec{y}\sin{\omega_{z}t}\right),$$
and convert to the rotating frame:
$$\mathbf{H_{1}} = H_{1}\left[\cos{\omega_{z}t}\left(\vec{x}'\cos{\omega_0t} - \vec{y}'\sin{\omega_{0}t}\right) + \sin{\omega_{z}t}\left(\vec{x}'\sin{\omega_0t} + \vec{y}'\cos{\omega_{0}t}\right)\right]$$
$$
\begin{gathered}
\mathbf{H_{1}} & = & H_{1}\left[\vec{x}'\left(\cos{\omega_0 t}\cdot\cos{\omega_{z}t} + \sin{\omega_{0}t}\sin{\omega_{z}t}\right) + \\ \quad \vec{y}'\left(\cos{\omega_{0}t}\sin\omega_{z}t - \cos{\omega_{z}t}\sin{\omega_{z}t}\right)\right]
\end{gathered}
$$
$$\mathbf{H_{1}} = H_{1}\left[\vec{x}'\cos{(\omega_{z}-\omega_{0})t} + \vec{y}'\sin{(\omega_{z}-\omega_{0})t}\right]$$
If you define $\Delta\omega = \omega_{z}-\omega_{0}$, then you have:
$$\mathbf{H_{1}} = H_{1}\left[\vec{x}'\cos{\Delta{\omega}t} + \vec{y}'\sin{\Delta\omega t}\right]$$
So if $\omega_{z}\rightarrow-\omega_{z}$, then the precession frequency is $-\left(\omega_{z}+\omega_{0}\right)$, which is $\Delta\omega-2\omega_{z}$, so you have:
$$\mathbf{H_{1}} = H_{1}\left[\vec{x}'\cos{(\Delta\omega - 2\omega_{z})t} + \vec{y}'\sin{(\Delta\omega - 2\omega_{z}) t}\right]$$
Which is to be expected, because you're mirroring the frequency about 0, not about $\omega_{0}$, so you pick up one factor of $\omega_{z}$ by going from $\omega_{z} \rightarrow 0$ and another factor of $\omega_{z}$ going from $0 \rightarrow -\omega_{z}$. Again, I'm a bit unclear on your notation, so if you were intending that $\omega_{z}$ is the frequency of precession of the rotating frame (what I'm calling $\omega_{0}$), then $\Delta\omega = 0$ and your signal will precess at $-2\omega_{0}$. You can see why from this diagram (the trace in blue is the on-resonance signal, the trace in green is the $-\omega_{0}$ signal):
The rotating frame is on resonance with $\omega_{0}$, and the signal is at $-\omega_{0}$, so in the rotating frame, the apparent frequency will double. One thing to note is that with some (maybe most?) NMR instruments, you're not actually going to be acquiring 2 channels of signal, so you're not actually getting a full quadrature signal. This is equivalent to looking at only the real channel of the signal. If you look at the Fourier transform of a real (or imaginary) channel only signal, you'll see that it's mirrored about 0, with negative and positive frequency components showing up on the spectrum:
So if $\omega_{z} = \omega_{0}$ and you acquire only one signal channel, then mirroring the field will have no effect in the rotating field.
If anyone wants it, here is the code I used to generate the above figures.
