Central characters are dot-Weyl orbits

If $\mu =w\cdot \lambda$ for some $w\in W$, then $\mu +\rho =w(\lambda +\rho )$. By Harish-Chandra isomorphism for the center, every central element $z$ determines a Weyl-invariant polynomial ${\mathrm{HC}}_{\rho }(z)\in S(\mathfrak{h}{)}^W$, and The Harish-Chandra projection computes the highest-weight scalar gives ${\chi }_{\lambda }(z)=\mathrm{pr}(z)(\lambda )={\mathrm{HC}}_{\rho }(z)(\lambda +\rho ),{\chi }_{\mu }(z)={\mathrm{HC}}_{\rho }(z)(\mu +\rho )\mathrm{.}$ Since ${\mathrm{HC}}_{\rho }(z)$ is ordinarily $W$-invariant, ${\mathrm{HC}}_{\rho }(z)(\mu +\rho )={\mathrm{HC}}_{\rho }(z)(\lambda +\rho )$. Hence ${\chi }_{\mu }(z)={\chi }_{\lambda }(z)$ for every central $z$, so ${\chi }_{\mu }={\chi }_{\lambda }$.

Conversely, if ${\chi }_{\mu }={\chi }_{\lambda }$, then the equalities from step 1.1 show that ${\mathrm{HC}}_{\rho }(z)(\lambda +\rho )={\mathrm{HC}}_{\rho }(z)(\mu +\rho )$ for every central $z$. Because Harish-Chandra isomorphism for the center identifies the image of ${\mathrm{HC}}_{\rho }$ with all of $S(\mathfrak{h}{)}^W$, every Weyl-invariant polynomial takes the same value at $\lambda +\rho$ and $\mu +\rho$. By [F1], those two points lie in the same ordinary $W$-orbit, so $\mu$ lies in the dot orbit of $\lambda$.

Therefore equal central characters are exactly the dot-Weyl orbits.