I have a question on Exercise 1.2.12 from Elizabeth Mansfield's book `A Practical Guide to the Invariant Calculus' (CUP, 2010).
Definition 1.2.11 Two group actions $\alpha_i:G\times M\rightarrow M$, $i=1,2$ are equivalent of there exists an invertible map $\phi:M\rightarrow M$ such that for all $g\in G$, $\alpha_2(g,z)=\phi^{-1}\alpha_1(g,\phi(z))$.
Exercise 1.2.12 Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be any invertible map, and define $\mu:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$ by $\mu(x,y)=f^{-1}(f(x)+f(y))$. Show $(\mathbb{R},\mu)$ is a group and thus defines an action of $\mathbb{R}$ on itself. Show that this action is equivalent to addition.
I could show the first part, and the identity element is then $e\in \mathbb{R}$ such that $f(e)=0$. But then to check the equivalence, let $\phi:\mathbb{R}\rightarrow \mathbb{R}$ be the sought invertible map. Then with $\alpha_2=\mu$, and $\alpha_1=+$, we must have $$ f^{-1}(f(g)+f(z))=\phi^{-1}(g+\phi(z)).\quad (\star) $$ In particular, with $g=e$, we get $$ z=f^{-1}(0+f(z))=f^{-1}(f(e)+f(z))=\phi^{-1}(e+\phi(z)), $$ i.e., $\phi(z)=e+\phi(z)$, giving $e=0$. So already it seems that there should have been the condition $f(0)=0$ given on $f$. But then even with this, with $z=e$ in ($\star$), we get $$ f^{-1}(f(g)+0)=\phi^{-1}(g+\phi(e)) $$ i.e., $g=\phi^{-1}(g+\phi(e))$, and so $\phi(g)= g+\phi(e)$. So $\phi$ is linear, $\phi(g)=g+k$, $k:=\phi(e)$ and $\phi^{-1}(g)=g-k$. Thus ($\star$) becomes $$ f^{-1}(f(g)+f(z))=(g+(z+k))-k=g+z, $$ i.e., $f(g)+f(z)=f(g+z)$, so $f$ is linear. Thus $\mu$ was addition to begin with. I am not sure where I am mistaken, or if there is something wrong with the claim in the exercise. I would be grateful for any help.