TITLE: Why are there $12$ automorphisms of $\Bbb Z\oplus \Bbb Z_{3}$?
QUESTION [10 upvotes]: Let $A:=\Bbb Z\oplus \Bbb Z_{3}$, then what is $|\text{Aut}(A)|$?  My answer is $4$ but the correct answer  (without explanation) turns out to be $12$! How come?
Well my understanding is, it just suffices to find out all the possibilities of $f(1,\bar 1)$ where $f$ is an arbitrary automorphism, since $(1,\bar 1)$ is the generator. So I think there are altogether $4$ possibilities: $(\pm 1, \pm \bar 1),(\pm 1, \mp \bar 1)$. How could there be any more?
I'd be very grateful if anyone would solve this puzzle for me! Thanks in advance.

REPLY [3 votes]: Consider presentation of $G$: $G=\langle x,y\colon y^3=1, xy=yx\rangle$.
Let $\sigma$ be any automorphism of $G$. Then $\sigma(y)$ could be $y$ or $y^2$ only (since $\langle y\rangle$ is torsion subgroup of $G$, so it is invariant under all the automorphisms). 
What can be $\sigma(x)$? Of course, it could be $x,x^{-1}$. Anything more? Yes. $xy$, $x^{-1}y$, $xy^2$, $x^{-1}y^2$. That's all. 
Thus, $\sigma(x)$ has two choices, $\sigma(y)$ has $6$ choices; each choice of $\sigma(x),\sigma(y)$ gives similar presentation of $G$, hence defines an automorphism. There are $12$.