Let $R=(R,+,\cdot)$ be a ring with $1_R$. We will write $U(R)=(U(R),\cdot)$ the multiplicative group of units of this ring and $R^+=(R,+)$ the additive group of this ring. The group $U(R)$ acts in the group $R^+$, if we define a group action $$*:U(R)\times R^+\longrightarrow R^+,\ (g,a)\mapsto g*a:=g\cdot a.$$ Then, we know that the function $$\phi: U(R)\longrightarrow \mathrm{Aut}(R^+),\ g\mapsto \phi(g):=\phi_g $$ is a group homomorphism. We can show that $\ker {\phi}=\{1_G\}$, so $\phi$ is injective and we have monomorphism.
My question: Is $\phi $ surjective?
Lets take an element $\psi \in \mathrm{Aut}(R^+)$. We want to find an element $g\in U(R)$ such that $\phi(g)=\psi$. Then $\phi _g=\psi \iff \phi_g(r)=\psi(r),\ \forall r\in R^+ \iff g\cdot r=\psi (r), \forall r\in R^+$.
But how can we solve the last equation, to find $g$? Any ideas please?
Thank you.