It appears this can happen already with the macro
average option. The statement needs some clarification, but I assume the precision and recall that are supposed to not bound the averaged F1 are themselves the same type of average.
Here's a simple example: $TP=TN=4$, $FP=1$, $FN=16$. Then
$$\begin{align*}
\operatorname{precision}(1)&=\frac{TP}{TP+FP}=0.8, \\
\operatorname{recall}(1)&=\frac{TP}{TP+FN}=0.2, \\
\operatorname{precision}(0)&=\frac{TN}{TN+FN}=0.2, \\
\operatorname{recall}(0)&=\frac{TN}{TN+FP}=0.8
\end{align*}$$
and so $F_1(1)=F_1(0)=0.32$, so the macro-average $F_1$ is also $0.32$. But the macro-averaged precision and recall are both $0.5$.