Let $z=a+bi$,

Since $i$ is the imaginary unit and the complex number $z$ satisfies $(1+i)z=1-i$,

Therefore, $(1+i)(a+bi)=a+ai+bi+bi^{2}=(a-b)+(a+b)i=1-i$,

Thus, $\begin{cases} a-b=1 \\ a+b=-1 \end{cases}$,

Solving this, we get $a=0$, $b=-1$,

Therefore, $z=-i$, $\overline{z}=i$.

Hence, the correct choice is: $\boxed{A}$.

By setting $z=a+bi$ and using $(1+i)(a+bi)=a+ai+bi+bi^{2}=(a-b)+(a+b)i=1-i$ to form a system of equations, we find $a=0$, $b=-1$, thus obtaining $z=-i$, $\overline{z}=i$.

This question examines the method of finding the conjugate of a complex number, the basic knowledge of algebraic operations of complex numbers, and the ideas of functions and equations. It is a basic question.