Firstly i'm not clear what the context is. So I'm assuming you're thinking of the additive group (Z/n) and you want to see when this is an injective homomorphism.
Okay, homomorphism first:
you want f(a + 
= f(a) + f(,
so 2(a + = 2a + 2b. So that's fine 
and f(0) = 2x0 = 0.
If it's injective the kernel must be trivial (Otherwise, suppose x non-trivial in the kernel. Then f(x) = f(0), so not injective).
Also trivial kernel => injective here. Prove this by contra-positive:
f not injective. So f(a) = f(, for some a not equal b, say.
then f(a - = f(a) - f( = f(a) -f(a) = 0. So a-b is in the kernel. Therefore the kernel is non-trivial.
So you're looking for all n such that the kernel is trivial. If you want to rule out a specific n, just find something non-trivial in the kernel and you're done.
I guess you've probably solved it by now but, if not, there's a bit of a nudge .
