From DFT theory, the Euler equation of an interacting system is:
$$\begin{align*} \frac{\delta E_{v}[\rho]}{\delta \rho(\vec{r})} &= \frac{\delta}{\delta \rho} \left[T_{s}[\rho]+\int \rho(\vec{r})v(\vec{r}) d \vec{r} +\frac{1}{2}\int\int \frac{\rho(\vec{r}) \rho(\vec{r}')}{|\vec{r}-\vec{r}'|}d \vec{r} d \vec{r}'+E_{xc}[\rho] \right] \\ &=\frac{\delta T_{s}[\rho]}{\delta \rho}+v(\vec{r})+\int \frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|} d \vec{r}'+\frac{E_{xc}[\rho]}{\delta \rho}=\mu \end{align*}$$
In Kohn-Sham approach, we introduce the auxiliary non-interacting system such that
$$ \frac{\delta E_{v_{s}}}{\delta \rho(\vec{r})}=\frac{\delta T_{s}[\rho]}{\delta \rho} + v_{s}(\vec{r}) = \mu $$
Then, we assume that 1st and 2nd equations are the same. In other words, two variation problems have the same solution, ground-state electron density $ \rho_{0}=\rho_{s0} $.
That's why we can get the ground-density from the auxiliary system i.e. $$\rho_{0} = \rho_{s0} = \sum_{i=1}^{occ} |\varphi_{i}(\vec{r})|^{2}$$
According to Hohenberg-Kohn theorem, an electron density $\rho$ has the one-to-one correspondence with a unique external potential $v[\rho]$.
However, $\rho_{0} = \rho_{s0}$ imply two different external potentials, or $v_{0} \neq v_{s}$.
I think that this would violates HK theorem. What mistakes did I made?