Let $\mathbb{D}=\{ z \in \mathbb{C} : |z|<1\}$ and its boundary $\mathbb{T}=\{ z \in \mathbb{C} : |z|=1\}$. If $\lambda \ne \mu$ are two points in $\mathbb{T}$, then I need to show that there is an automorphism $v$ of $\mathbb{D}$ such that $v(\lambda)=1$ and $v(\mu)=-1$.
I tried the following : Any automorphism of $\mathbb{D}$ is of the form $v(z)=\eta \frac{z-a}{1-\overline{a}z}$ for some $\eta \in \mathbb{T}$ and $a \in \mathbb{D}$. If we put $v(\lambda)=1$ and $v(\mu)=-1$, we get the following:
$\eta (\lambda-a)= 1-\overline{a}\lambda \quad \text{and} \quad \eta (\mu-a)= -1+\overline{a}\mu$.
If we try to solve for the value of $a$, we get that $(\lambda-a)(-1+\overline{a}\mu)= (1-\overline{a}\lambda)(\mu-a)$. But I have not been able to solve this equation.
By rotating the initial configuration we can assume that $\lambda$ and $\mu$ are symmetric with respect to the real axis and lie in the closed right halfplane, i.e. $$ \lambda = x+ iy \, , \mu = x-iy \, , \quad 0 \le x < 1 \, , 0 < y \le 1 \, . $$ The equation $$ (\lambda-a)(-1+\overline{a}\mu)= (1-\overline{a}\lambda)(\mu-a) $$ then becomes $$ 2x (1+|a|^2) + 4 \operatorname{Re}(a) = 0 \, . $$ The symmetry of the rotated configuration suggests to look for a real solution $a$, which then must satisfy the equation $$ \tag{$*$} x (1+a^2 ) + 2a = 0 \, . $$ If $x = 0$ then $\boxed{a=0}$ is a solution. Otherwise $(*)$ is a quadratic equation in $a$ with the solutions $$ a = \frac{1 \pm\sqrt{1-x^2}}{x} = \frac{1 \pm y}{x} \, . $$ Exactly one of these solutions, namely $\boxed{a = \frac{1-y}{x}}$ is in the range $(0, 1)$.
So on any case we have found a real value $a \in [0, 1)$ such that the corresponding Möbius transformation $T(z) = \frac{z-a}{1-az}$ satisfies $T(\lambda) = - T(\mu)$. It only remains to append a rotation with a factor $\eta$ of modules one to get $v(\lambda)=1$ and $v(\mu)=-1$.
There is also a geometric solution: Find $a \in \Bbb D$ such that the four points $\lambda, a, \mu, 1/\bar a$ lie on a circle $C$. $T(z) = \frac{z-a}{1-\bar az}$ maps $C$ onto a line $L$ through the origin. Then $T(\lambda)$ and $T(\mu)$ lie on the boundary of the unit disk and on $L$, so that $T(\lambda) = - T(\mu)$. Again it only remains to compose that with a rotation of unit modulus.
Yet another approach: Choose $\nu, \zeta \in \Bbb T$ such that $\nu \ne \lambda, \tau$ and $\zeta \ne 1, -1$. There is a unique Möbius transformation $T$ which maps $\lambda, \mu, \nu$ to $1, -1, \zeta$, respectively. $T$ maps $\Bbb T$ only itself, so that either $T$ or $1/T$ is an automorphism of the unit disk with the desired properties.