Let $X_1,\cdots,X_{n}$ be i.i.d.r.v.'s from $N(\theta,1),$ where $\theta$ is unknown.Show the statistic $T(\mathbf{X})=\sum_{i=1}^{n},X_{i}/n=\bar{X} $ is sufficient for $\theta$.

The answer as following:

Suppose that we transform variables $X_1,\cdots,X_{n}$ to $Y_1,\cdots,Y_{n}$ with the help of an orthogonal transformation so that $Y_{1}$ is $N(\sqrt{n}\theta,1),Y_{2},\cdots,Y_{n}$ are i.i.d. $N(\theta,1)$,and $Y_{1},\cdots,Y_{n}$ are independent. We take $$y_{1}=\sqrt{n}\bar{x},$$ and ,for $$k=2,\cdots,n,\quad y_{k}=\left [(k-1)x_{k}-(x_1+\cdot+x_{k-1}) \right ]/\sqrt{k(k-1)}.$$

And then, the author said that "$\bar{X}$ with respect to $(X_1,\cdots,X_{n})$ is sufficient for $\theta$ $\Leftrightarrow $ $Y_{1}$ with respect to $(Y_1,\cdots,Y_{n})$ is sufficient for $\theta.$" I know that $Y_{1}$ with respect to $(Y_1,\cdots,Y_{n})$ is sufficient for $\theta.$ but why $\bar{X}$ is sufficient for $\theta$ is equivalent to $Y_{1}$ is sufficient for $\theta$? How to prove this rigorously?