Group of units of End($\mathbb{Z}/_2 \oplus \mathbb{Z}/_2) \cong S_3$? Consider the monoid End($\mathbb{Z}/_2 \oplus \mathbb{Z}/_2$) of endomorphisms of the abelian group $\mathbb{Z}/_2 \oplus \mathbb{Z}/_2$ under composition. Show that the group of units of End($\mathbb{Z}/_2 \oplus \mathbb{Z}/_2)$ is isomorphic to the symmetric group $S_3$.
So far, I have:
The group of units of End($\mathbb{Z}/_2 \oplus \mathbb{Z}/_2$) contains all of the endomorphisms that are invertible. These are the endomorphims that do not map components to 0; thus, there are 6 elements in the group of units: 
    \begin{align*}
 \rho_1: & (x,y) \mapsto (x,y) & \rho_2: & (x,y) \mapsto (y,x)\\
 \rho_3: & (x,y) \mapsto (x, x+y) & \rho_4: & (x,y) \mapsto (x+y, x) \\
 \rho_5: & (x,y) \mapsto (y, x+y) & \rho_6: & (x,y) \mapsto (x+y, y) 
 \end{align*}
On the other hand, the symmetric group $S_3$ also has 6 elements: 
    \begin{align*}
 (1)(2)(3) & & (12)(3) & & (13)(2) & & (23)(1) & & (123) & & (321).
 \end{align*}
To show the isomorphism, I just need to exhibit a bijection. What do I need to do to define a bijection between these two groups? I can map the identity to the identity, so $\rho_1 \mapsto (1)(2)(3)$, but then can I just arbitrarily pair off the rest of the elements?

A unit in the ring of endomorphisms of an abelian group is an automorphism. The automorphisms of the Klein four group permute the three nontrivial elements, and are determined by such.
