How to show that $N(\varphi (a) ) = \varphi (N(a))$, where $N(b) = \{ x \in G | xb = bx \}$ is the normalizer of $b$ in $G$?

Question

Let $G$ be a group, let $a \in G$ and $ \varphi$ an automorphism of $G$.

Show that $N(\varphi (a) ) = \varphi (N(a))$, where $N(b) = \{ x \in G | xb = bx \}$ is the normalizer of $b$ in $G$

Answer 1

Observe that (using $\;C(a)\;$ for the centralizer of the element $\;a\;$, instead of the probably misleading $\;N(a)\;$

$$x\in C(\phi(a))\iff x\phi(a)=\phi(a)x$$

But $\;x=\phi(y)\;$ , for some $\;y\in G\;$ , so the above can be written as

$$x\in C(\phi(a))\iff x\phi(a)=\phi(a)x\iff\phi(y)\phi(a)=\phi(a)\phi(y)\iff$$

$$\iff\phi(ya)=\phi(ay)\iff x=\phi(y)\in\phi(C(a))\,,\,\,\text{since clearly}\;\;y\in\ C(a) .$$