Let group $G=\{1, a, b, x, y, z\}$ and $\alpha$ be an automorphism of $G$ such that $$\alpha(1)=1;$$ $$\alpha(a)=b, \alpha(b)=a;$$ $$\alpha(x)=y, \alpha(y)=z, \alpha(z)=x $$
My question is:
\begin{align} \alpha(\alpha(a))=\alpha(b)=b\text{ gives } \alpha^2=1 \text{ i.e. it is an automorphism of order } 2; \end{align}
whereas$$\alpha(\alpha(\alpha(x)))=x \text{ gives } \alpha^3=1 \text{ i.e. it is an automorphism of order } 3. $$
It is a contradicting conclusion. What have I missed? Any help would be appreciated!
You only get that $\alpha^2(a)=a$ and $\alpha^2(b)=b$. Similarly, for $t\in\{x,y,z\}$, $\alpha^3(t)=t$.
This is not a contradiction.
Note that, for every automorphism $\beta$ of a group $\Gamma$, $$ \{x\in\Gamma:\beta(x)=x\} $$ is a subgroup of $\Gamma$.
Can you finish? Hint: $\alpha$ cannot exist.