Verified answer

Z_2 ⊕ Z_2 only has four elements, so you can easily write out all possible bijective maps and determine which are isomorphisms. Actually, you know any isomorphism must take (0,0) to itself, so that greatly reduces the possibilities. In this case it turns out in fact that any bijective map from Z_2 ⊕ Z_2 to itself that fixes (0,0) is an automorphism. Here are the possibilities:

Automorphism 1 (identity):

(0,0) -> (0,0)

(1,0) -> (1,0)

(0,1) -> (0,1)

(1,1) -> (1,1)

Automorphism 2:

(0,0) -> (0,0)

(1,0) -> (0,1)

(0,1) -> (1,0)

(1,1) -> (1,1)

Automorphism 3:

(0,0) -> (0,0)

(1,0) -> (1,1)

(0,1) -> (0,1)

(1,1) -> (1,0)

Automorphism 4:

(0,0) -> (0,0)

(1,0) -> (1,0)

(0,1) -> (1,1)

(1,1) -> (0,1)

Automorphism 5:

(0,0) -> (0,0)

(1,0) -> (0,1)

(0,1) -> (1,1)

(1,1) -> (1,0)

Automorphism 6:

(0,0) -> (0,0)

(1,0) -> (1,1)

(0,1) -> (1,0)

(1,1) -> (0,1)

As you can see, Aut(Z_2 ⊕ Z_2) is isomorphic to S_3, through the action of S_3 that permutes the nonidentity elements.