Let $Q$ be the quaternion group of order $8$. Write $G = \operatorname{Aut}(Q)$ and $N = \operatorname{Inn}(Q)$. Show that $C_G(N) = N$
What I tried
I know that $C_G(N) = \{g \in G : gN = Ng\}$. I already know that $nN =Nn$ if $n \in N$, which means that $C_G(N) \supseteq N$.
For $C_G(N) \subseteq N$, all I need to show is that $$ \forall g \in G \setminus N, \exists n \in N, \quad gng^{-1} \notin N $$ In other words, I have to show that if $\phi : Q \rightarrow Q$ is an arbitrary isomorfism, that cannot be written as a conjugation, there is some $y \in Q$ such that $$ \text{for each }q \in Q, \qquad \left(x \ \mapsto \ \phi(y\phi^{-1}(x)y^{-1})\right) \quad \neq \quad \left( x \ \mapsto \ qzq^{-1}\right) $$ I rewrote $\phi(y\phi^{-1}(x)y^{-1}) = \phi(y)x\phi(y)^{-1}$
But this is a conjugation right? Did I mess things up somewhere?