Is the PR AUC invariant under label flip?
I propose to look at this a bit differently. The different classification status of the two classes are related as follows:

$TP_1=TN_0$
$FP_1=FN_0$
$FN_1=FP_0$
$TN_1=TP_0$

From there we have:
$$FPR_0=\frac{FP_0}{FP_0+TN_0}=\frac{FN_1}{FN_1+TP_1}=1-\frac{TP_1}{FN_1+TP_1}=1-TPR_1$$
and naturally $FPR_1=1-TPR_0$ for the same reasons. Therefore when one switches the labels on a ROC curve, one obtains a mirror image of the curve using the top-left to bottom-right diagonal as the symmetry axis. It follows that the AUC is identical.
However this does not hold with a PR curve, because there is no such symmetry between precision and recall of the two classes. A counter-example should be enough to prove this:
Suppose a test set with 100 instances, 90/10 of which have true class 0/1. Let's assume the classifier predicts all the instances as class 0 except 2 instances as class 1, one true 0 and one true 1.
$$
\begin{array}{|c|c|c|}
\hline
&amp;predicted~0 &amp; predicted~1 \\
\hline
\hline
true~0 &amp; 89 &amp; 1 \\
\hline
true~1 &amp; 9 &amp; 1 \\
\hline
\end{array}
$$
Thus we have:
$P_0=0.92, R_0=0.99, P_1=0.5, R_1=0.1$
[edited] Assuming there is only this point on the PR curve and the two following extreme cases:

Predict every instance as class 0: $P_0=0.9, R_0=1, P_1=NaN, R_1=0$
Predict every instance as class 1: $P_0=NaN, R_0=0, P_1=0.1, R_1=1$

Note: I'm not totally sure about the proper way to deal with $P=NaN$ but I think it's ok to assume P=1 in order to put these points on the curve (this is not essential to the point anyway).
It's easy to see that the AUC is very different whether the focus class is 0 or 1:

In the former case the curve is entirely in the top part of the graph and the area is almost the maximum.
In the latter the curve is mostly in the bottom part of the graph and the area is less than 0.5.