At $n=2$ the sample mean and the midrange are literally the same statistic

Let $X_1,X_2,\dots,X_n$ be iid $U(0,\theta)$. Define $T_1=\bar X$ and $T_2=(X_{(1)}+X_{(n)})/2$, where $X_{(1)}=\min(X_1,\dots,X_n),\ X_{(n)}=\max(X_1,\dots,X_n)$. Then: