We can identify the group of automorphisms of the upper half plane with $PSL(2,\mathbb{R})$, the group of $2\times 2$ matrices with determinant 1, up to $\pm I$. That is, they are transformations of the form
$$z \rightarrow \frac{az + b}{cz + d},$$
where $ad - bc = 1$ and $(a,b,c,d) \sim (\alpha a, \alpha b, \alpha c, \alpha d)$ for $\alpha \neq 0$, $a, b, c, d \in \mathbb{R}$. Observe that this is a three-parameter family.
If there were two such automorphisms $\Phi_1$ and $\Phi_2$, then $\Phi_1^{-1}\circ\Phi_2(x_j) = x_j$ for $j = 1, 2, 3$, i.e. there are three fixed points. The only automorphism which has three fixed points is the identity, so $\Phi_1^{-1}\circ\Phi_2 = I$. (Why? If $(az+b)/(cz+d) = z$ is a fixed point, then unless $c = a = 1$ and $b = d = 0$, we obtain at best a quadratic with at most two roots, by the fundamental theorem of algebra.)
Recall Poisson's integral formula for the half-plane says that if $U : \mathbb{R} \to \mathbb{R}$ is piecewise continuous and bounded, then
$$
f(x,y) = \frac{y}{\pi} \int_{-\infty}^{\infty} \frac{U(t)}{(x-t)^{2} +y^{2}} dt,
$$
is harmonic in the upper half-plane and has boundary values $f(x,0) = U(x)$.
So, given your boundary values, we have
$$
U(x) = \begin{cases} -1 & x < 0 \\ 1 & x > 0. \end{cases}
$$
Note, as we are integrating, we don't care what the value of $U(0)$ is.
Then,
\begin{equation}
f(x,y) = \frac{y}{\pi} \int_{-\infty}^{\infty} \frac{U(t) dt}{(x-t)^{2} +y^{2}} = \frac{y}{\pi} \left[ \int_{0}^{\infty} \frac{dt}{(x-t)^{2} + y^{2}} - \int_{-\infty}^{0} \frac{dt}{(x-t)^{2} + y^{2}}\right].
\end{equation}
I'll compute the first integral, and leave the second as practice. Using the change of variables $u = x-t$, we see that $d u = -dt$. So,
\begin{align*}
\int_{0}^{\infty} \frac{dt}{(x-t)^{2} + y^{2}} &= \int_{0}^{\infty} \frac{-du}{u^{2} + y^{2}}\\
& = \frac{-1}{y} \arctan\left( \frac{x-t}{y} \right) \bigg|_{t=0}^{\infty}\\
& = \frac{1}{y} \arctan\left(\frac{t-x}{y}\right) \bigg|_{t=0}^{\infty} \\
& = \frac{1}{y} \left( \frac{\pi}{2} - \arctan \frac{x}{y} \right).
\end{align*}
Now use similar technique to perform the other integral. Don't forget the $\frac{y}{\pi}$ out in front of Poisson's formula, and be wary of negative signs.
Best Answer
Developing @MoisheKohan's hint, we get that $cz^2+(d-a)z-b$ is the zero polynomial. Thus $b=c=0$ and $a=d $.
So $\phi (z)=z $.