Prove this is an automorphism Let $r\in U(n)$. Prove that the mapping $\phi:Z_n \rightarrow Z_n$ defined by $\phi(s)=sr$ mod$n$ $\forall s \in Z$ is an automorphism of $Z_n$.
My first question is $U(n)$ means $U_n$ right? I have not seen it this way ever.
Secondly, to prove it is a bijection, assume $\phi(x)=\phi(y)$. So $xr\equiv yr \mod n$. Thus $r(x-y)\equiv0 \mod n$. What is next? 
Then for the onto part, we must show that $p\in\mathbb{Z}_n$ implies that there exists a $q\in\mathbb{Z}_n$ such that $\phi(q)=qr\equiv p\mod n$. How do I show this?
Finally I have to show that it preserves operation. So we must show $\phi(ab)\equiv\phi(a)\phi(b)\mod n$. 
Q.E.D.
If someone could point me in the right direction or show that I am incorrect that would be great.
 A: To establish bijectivity, we need only verify injectivity as the domain and codomain of $f_r$ have the same (finite) cardinality.  As $r\in U(n)$, there is a unique inverse $r^{-1}\in\mathbb{Z}_n$.  Now for all $x,y\in\mathbb{Z}_n$ we have
$$f_r(x)=f_r(y) \iff xr=yr \iff x=xrr^{-1}=yrr^{-1}=y$$
Thus $f_r$ is injective, and by our comments above it follows that $f_r$ is a bijection.
To see that $f_r$ is a homomorphism (on addition), let $x,y\in\mathbb{Z_n}$.  Now we have that $f_r(x+y)=(x+y)\cdot r=x\cdot r+y\cdot r=f_r(x)+f_r(y)$ as desired.
Hence $f_r$ is an automorphism of $\mathbb{Z}_n$ under addition as desired.
A: Hint: the key observation is the same for both 1-1 and onto: what does $r\in U(n)$ mean? What property does $r$ have modulo $n$? (Hint-within-a-hint: it allows you to "cancel" the $r$ in the 1-1 proof).
A: Your proof looks good the reason you need r $\in$ U(n) is that you know r invertible so suppose p $\in$ $Z_n$ since r $\in$ U(n) There exists $r^{-1}$ which exists in $Z_n$ by assumption so this multiplication $pr^{-1}$ $\in$ $Z_n$ is well defined so $\phi(pr^{-1}) = p$ and you are done. Also for your injective part you can multiply by inverse in both side since you know $r$ $\in$ U(n) that is the set of invertible elements of $Z_n$.
Now to show that is an homomorphism that is trivial since $\phi(p_1 + p_2) = (p_1 + p_2)r = p_1r + p_2r = \phi(p_1) + \phi(p_2)$ and then that proves that this is indeed a automorphism.
