When a function of sufficient statistic is itself sufficient? I'm following notes at onlinecourses and I got confused on transformation of sufficient statistics. For example, if $X$ is a sufficient statistic for $\mu$, why $Y=X^2$ is not a sufficient statistic for $\mu$? In another source Essentials of Statistical Inference by G. A. Young p. 92 $|X|$ is sufficient. Aren't both $X^2$ and $|X|$ many-to-one functions? Why in Young's book the sign no longer matters? Is this something to do with having single observation? Apologies in advance if this question was already asked.
Extract from onlinecourses:

Later in the text the following is mentioned:

Here is an extract from Young's book:

 A: This depends on context. If $X$ (possibly a vector) is an observation from some statistical model, and $T=T(X)$ is sufficient, then any one-to-one function of $T$ is also sufficient,  see Function of a sufficient statistic.  But in some cases a function of $T$ which is not one-to-one might also be sufficient. In such cases $T$ will not be minimal sufficient. So on to your examples:


*

*Say $X=(X_1, \dotsc, X_n)$ is a sample from the model $\mathcal{Norm}(\mu, 1)$. Then $T=\bar{X}_n$ is sufficient, and so is any one-to-one function of $T$, like $T^{1/3}$. But the square is not one-to-one, so say if $T=1000$, then if you report only $T^2=1000000$, you loose the information about the sign, so cannot discriminate between the models $\mu=1000$ and $\mu=-1000$. On the basis of $T=1000$ you can discriminate those models. That should explain why $T^2$ in this model is not sufficient.  

*The other example, the model $\mathcal{Norm}(0, \sigma^2)$. This model is a symmetric distribution, and observing $T=1000$ or observing $T=-1000$ gives the same information about the variance $\sigma^2$. So in this case $\vert T \vert$ or $T^2$ are still sufficient. 
