I am working on an example of the Drinfeld Double of the Group Algebra and stumbled upon the book On Characters of Finite Groups. My issue is with 8.1.1, page 192 of the PDF (relevant part here). It says:
The Group $G$ acts on $\operatorname{Hom}(G,K) = \operatorname{Fun}(G,K)$ the $K$-valued maps from $G$ with conjugation: For $g \in G$ and $\varphi \in \operatorname{Fun}(G,K)$ we set $g.\varphi := g \varphi g^{-1}$, that is $(g.\varphi)(s) := \varphi(g^{-1}sg)$ for all $s \in G$.
For me it seems that there are two different definitions for the action which gets me confused ($(g.\varphi)(s) := \varphi(gsg^{-1})$ and $(g.\varphi)(s) := \varphi(g^{-1}sg)$). I do not think that it is an error in the book as the author states $(g.\varphi)(s) := \varphi(g^{-1}sg)$ again somewhere else.
What is meant with $g.\varphi := g \varphi g^{-1}$?
If I try to check that $(g.\varphi)(s) := \varphi(g^{-1}sg)$ is an action, I encounter a problem:
\begin{align*} (g.(h.\varphi))(x) &= g. \varphi (h^{-1}xh) = \varphi ( g^{-1}h^{-1}xhg) \\ &= \varphi ((hg)^{-1}xhg) = (hg).\varphi (x) \end{align*} which is not $(gh).\varphi(x)$ for all $x \in H$.
Just a guess, but $\operatorname{Fun}(G,K)$ can be endowed with commuting standard left and right translation actions given by $(g\cdot \varphi)(x)=\varphi(xg)$ and $(\varphi\cdot g)(x)=\varphi(gx)$. Then if these actions are implicit in the notation $g\varphi g^{-1}$, you get the definition Broué gives.
It is indeed an action since $$ (g.(h.\varphi))(x)=(h.\varphi)(g^{-1}xg)=\varphi(h^{-1}g^{-1}xgh)=\varphi((gh)^{-1}x(gh))=((gh).\varphi)(x). $$
In the first equality, $g$ is acting on the function $h.\varphi$ first, which is why your calculation came out backwards.