Okay, so at first I was with you and thought for sure this function was impossible, and I am still not completely convinced this works, but here is what I am thinking.

First the motivation.

f(n)=a for some a

then you know the value of f(a) to satisfy the requirement

f(a)=-n

f(-n)=-a

f(-a)=n

but then this sort of cycle of 4 works to fit the definition. I have built these ones to satisfy the requirements. So basically you just gotta find a nice way to partition the integers into dijoint cycles of 4 so this will work out and here is what I came up with.

f(0)=0

f(1)=2

f(2)=-1

f(-1)=-2

f(-2)=1

f(3)=4

f(4)=-3

f(-3)=-4

f(-4)=3

.

.

.

This is clearly bijective, so it makes sense definition wise.

so basically I think the closed form is

f(0)=0

for n positive and odd f(n)=n+1

for n positive and even f(n)=-n+1

for n negative and odd f(n)=n-1

for n negative and even f(n)=-n-1

Okay, I think it works, see what you think.