Suppose $Q_8\le G$ where $G$ has an element of order $3$, call it $a$.
Let $\psi:Q_8\longrightarrow Q_8$ be defined by the conjugation by $a$, i.e. $\psi:q\mapsto q^a$.
Suppose to know that this is not a trivial map and that $(Q_8)^a=Q_8$.
I have to prove that $\psi\in\operatorname{Aut}(Q_8)$.
- $\psi$ is easily a group omomorphism.
- $\psi$ is not trivial and surjective (since $(Q_8)^a=Q_8$) by the hypotesis.
- $\ker\psi=C_{Q_8}(a)$ but I'm not able to show that it's trivial. It should, but seeing $Q_8$ as a set of square matrixes of order $2$, we note that $Z(Q_8)=\{\pm\mathbb I\}$ from which $\psi$ wouldn't be injective.
Where is the problem in the third point?
Thank you all
Kernel is not center in that case,
remember that $$ker=\{x\in Q_8 | x^a=e\}$$
But if $x^a=e$ then $x=e$.
I think you are confusing the terms fixing the elements and sendinig the elements to identity.