How does the conversion between admittance and impedance affect phase angle?
You can analyze phasor notation in a quasi-2D Cartesian fashion. The real part is the "x", and the complex part is the "y".

So given a phasor magnitude M with angle Theta,

Using trig:
\begin{equation}
R = M \cos(\theta)\\
X = M \sin(\theta)
\end{equation}

We now have the complex impedance R + Xj

To invert, you can multiply by the complex conjugate (R - Xj) to both the numerator and denominator.

\begin{equation}
Y = \frac{R - Xj}{(R + Xj)(R - Xj)} = \frac{R - Xj}{R^2 + X^2}
\end{equation}

To compute the magnitude of the admittance, use the distance formula:

\begin{equation}
M_Y = \sqrt{\left(\frac{R}{R^2 + X^2}\right)^2 + \left(\frac{-X}{R^2 + X^2}\right)^2}
\end{equation}

And the phase of the admittance:

\begin{equation}
\theta_Y = \tan^{-1}\left(\frac{-X}{R}\right)
\end{equation}

Note that tangent is a bit finicky for computing the phasor angle as you have to be careful about the quadrant. If you're using a computer, they often times have an "atan2" function which takes the x and y coordinates directly and computes the CCW angle from the positive X axis.

A closer look at the phase angle mapping, and it looks like the admittance phase angle is just the reflection of the impedance phase angle about the real/X axis.

For example, an impedance phase angle of 45 degrees is equal to an admittance phase angle of -45 degrees.

And this makes sense if I had used some identities above:

\begin{equation}
\theta_Y = -\tan^{-1}\left(\frac{X}{R}\right) = -\theta_X
\end{equation}