Given an $(x, y, z)$ system I create a new system $(x', y', z')$ by applying two rotations $\theta$ and $\phi$.
In the new system the $(x',y')$ plane, i.e.: the $z'=0$ plane, can be written as:
$$ (sin\theta\,sin\phi)\,x + (-cos\theta\,sin\phi)\,y + (-cos\phi)\,z = 0$$
I need to find out the smallest range of $\theta$ and $\phi$, necessary to produce all possible configurations of the $(x',y')$ plane.
For example, the angles $(\theta=45, \phi=26)$ will produce the same plane as $(\theta=45, \phi=206)$.
One range that would work is $\left\{0<\phi\le90,0\le\theta<360\right\}\bigcup\left\{\phi=0,0\le\theta<180\right\}$.The reason why this range works is it restricts one of the unit normals of your plane to the hemisphere above the $xy$-plane. A plane through the origin is uniquely determined by its pair of unit normals. You can think of $\phi$ as the angle of the unit normal from the positive $z$ axis, and $\theta$ as a rotation about the $z$ axis. In the special case where the normal lies in the $xy$-plane, we take care to limit it to half the unit circle in the $xy$-plane.
Edit:
Found a bug in my original suggestion when examinging Gabriel's range (see comment below): $\left\{-90<\phi\le90,0\le\theta<180\right\}$. Thinking about this geometrically made me feel it was correct, but I couldn't articulate a watertight geometric argument so came up with the following algebraic argument.
The criteria for uniqueness of representation over a given range can be written as $$\begin{bmatrix}\sin\theta_1\sin\phi_1\\\cos\theta_1\sin\phi_1\\\cos\phi_1\end{bmatrix}=k\begin{bmatrix}\sin\theta_2\sin\phi_2\\\cos\theta_2\sin\phi_2\\\cos\phi_2\end{bmatrix},k=\pm1\text{ implies }\theta_1=\theta_2\text{ and }\phi_1=\phi_2$$ (In words, two data points for $\theta$ and $\phi$ giving equivalent normals implies it's the same data point)
With $-90<\phi\le90$ we have $0\le\cos\phi\le1, -1 < \sin\phi\le1$.
With $0\le\theta<180$ we have $-1<\cos\theta\le1,0\le\sin\theta\le1$
Case $k=-1$: $\cos\phi_1=-\cos\phi_2$ implies $\cos\phi_1=\cos\phi_2=0$, so $\phi_1=\phi_2=90$. Then the other equations become $\sin\theta_1=\sin\theta_2$ and $\cos\theta_1=\cos\theta_2$ implying $\theta_1=\theta_2$ as required.
Case $k=1$: $\cos\phi_1=\cos\phi_2$ implies $\phi_1=\pm\phi_2$. This rules out $\phi_i=90$.
Subcase $\phi_i\ne0, \phi_1=\phi_2$: $\sin\phi_i\ne0$ so the remaining equations become $\sin\theta_1=\sin\theta_2$ and $\cos\theta_1=\cos\theta_2$ implying $\theta_1=\theta_2$ as required.
Subcase $\phi_i\ne0, \phi_1=-\phi_2$: $\sin\phi_i\ne0$ and the remaining equations become $\sin\theta_1=-\sin\theta_2$ and $\cos\theta_1=-\cos\theta_2$ which due to the range of $\theta$ implies $\theta_1=\theta_2=0$, but then $\cos\theta_0=-\cos\theta_2$ implies $1=-1$, so there is no such valid pair $\phi_1,\phi_2$.
Subcase $\phi_1=\phi_2=0$. This is where it all comes unstuck. $\sin\phi_1=\sin\phi_2=0$ and the other equations become $0=0$, meaning any valid values of $\theta_1$ and $\theta_2$ will work.
So we can correct the suggested valid ranges to: