Is independence preserved under transformation? If not, assume linear/continuous?
This problem comes from proving $X_n$ and $s_n^2$ are independent given iid sample from $N(\theta,\sigma^2)$
The argument goes $s_n^2$ is a function of $X_i-\bar{X}$ which is uncorrelated to $\bar{X}$ by Basu's theorem.
Yes it is.
If $f:(\mathcal{X},\mathcal{A})\rightarrow(F,\mathcal{F})$ and $g:(\mathcal{Y},\mathcal{B})\rightarrow(G,\mathcal{G})$ are measurable transformations and $X,Y$ are independent random variables taking respective values in $(\mathcal{X},\mathcal{A}),(\mathcal{Y},\mathcal{B})$, then $f(X),g(Y)$ are also independent. This follows from the fact that $$\sigma(f(X))=\{X^{-1}[f^{-1}(F)]:F\in\mathcal{F}\}\subset\sigma(X^{-1} A:A\in\mathcal{A})=\sigma(X),$$ from an analogous calculation $\sigma(g(Y))\subset\sigma(Y)$, and from the definition of independence.
Basu's theorem, in one of it's implications states that if $T$ is a sufficient and bounded complete statistic, and $U$ is an ancillary statistic then $T$ and $U$ are independent. In the particular case in question $T(X)=\bar X$ is known to be (minimal) sufficient and complete. Further, $$U(X)=(X_1-\bar X,\dots,X_n-\bar X)=(Z_1-\bar Z,\dots,Z_n-\bar Z),$$ where $Z_i:=X_i-\theta$ does not depend on $\theta$. Hence $U$ is ancillary $\textbf{if $\sigma^2$ is a known parameter}$. In that case, by the above result, even $\bar X$ and $s^2=s_n^2=||U||_2/(n-1)$ are independent.
Note that $X_n$ and $s_n^2$ in general need not be independent.