Let $(G,*)$ be a finite group and $\psi \in \text{Aut}(G)$ a fixedpoint-free automorphism, that is, if $\psi(x) = x$, then $x = 1$.
In a proof I am reading there is the following incidental remark: If $g \in G$, then the map $x \mapsto g * \psi(x) * g^{-1}$ is "clearly" again a fixedpoint-free automorphism.
Question: Why ist this automorphism fixedpoint-free? It is not obvious for me.
Thank you for your thoughts!
EDIT: It follows the original proof mentioned above: the author uses postfix notation for maps, i.e., $x\psi$ for $\psi(x)$. All rules are formed accordingly.
EDIT2: Is the proof of the following really "obvious"?
Okay, consider this reference, in particular page $17$ of the document $(26$ of the pdf$)$, for the proof of Lemma $5.2$. In the reference, it's on letter $(d)$ of Lemma $3.2$.
Now, the reference also answers the main question. Indeed, let $x$ be a fixed point of $\psi_g$. Then $x=$ $g\,\psi(x)\,g^{-1}$, that is $x$ and $\psi(x)$ are conjugate, by $g$. As shown in letter $(c)$ of Lemma $3.2$ in our reference, this implies $x=1_G$. In other words, the only fixed point of $\psi_g$ is the identity, and hence $\psi_g$ is fixed-point-free.
As a side note, your Lemma $5.2$ implies, by direct computation, that $\psi_g$ also satisfies $\psi_g^3=\text{Id}$.