Let $A$ be a $k$-algebra, $k$ a field of characteristic zero. Let $g$ be a $k$-algebra automorphism of $A$, and let $g^{-1}$ be its inverse.
Assume that $g^{-1}=u^{-1}gu$, for some automorphism (or anti-automorphism) $u$.
Is being conjugate implies something 'interesting' about $g$ and $g^{-1}$?
I am not sure what 'interesting' exactly means; for example, $g$ and $g^{-1}$ have the same order (this is the only interesting thing I have succeeded to say thus far).
Does it matter if $A$ is commutative or not?
A special case: Actually, I have a non-commutative algebra $A$ in mind, the first Weyl algebra $A_1$ (over $\mathbb{C}$, for example):
Denote the group of automorphisms of $A_1$ by $G$, and the group of automorphisms and anti-automorphisms of $A_1$ by $H$. Let $\beta: (x,y) \mapsto (x,-y)$. Then $H=G \cup G\beta$. Let $f$ be an involution on $A_1$ (= an anti-automorphism of order two). $f=g\beta$, for some automorphism $g$, and $f^2=1$ means $g \beta g \beta =1$, so $g^{-1}= \beta g \beta$ ($u=\beta$). I wonder if the order of $g$ in this case must be finite (probably not?).
Thank you very much!