I came across this problem on brilliant. and to my understanding, the definition of automorphisms is to map a group to itself with bijective mapping which means the origin group and the mapped group must have elementwise pairings.
and according to brilliant answer:
An automorphism is determined uniquely by what elements the three generators $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ turn into under the map. The first one can turn into any of seven nontrivial elements. The second one can turn into any of six elements (not the identity and not what $(1,0,0)$ turned into). The third one can turn into any of four elements (it can't turn into any of the four elements generated by the images of the first two generators). So, the answer is $7 \cdot 6 \cdot 4 = 168$.
In fact, the automorphism group is $GL_3({\mathbb F}_2)$, the group of with coefficients in $\{0, 1\}$ modulo $2$.
I wonder how could it be possible to let's say map
(1, 0, 0) to (0, 0, 1),
(0, 1, 0) to (0, 1, 0) and
(0, 0, 1) to (0, 1, 1) and still have (1, x, x) mapped back to origin group? the first 1 seems not to be able to be generated through any combinations of mapped generators.
Your counterexample does not satisfy the conditions required by the construction.
Starting with $$ (1,0,0) \mapsto (0,0,1) $$ is fine, as is extending the construction by $$ (0,1,0) \mapsto (0,1,0) $$ But the construction then requires that $(0,0,1)$ is not sent to any element that can be generated by the images of the generators considered so far.
However, $(0,0,1)+(0,1,0) = (0,1,1)$, so the conditions prohibit sending $(0,0,1)$ to $(0,1,1)$.