Let $e_1$, $e_2$ be an orthonormal basis of $T_{p}S$. Let:
\begin{align*}
\langle d\phi_{p}(e_1), d\phi_{p}(e_1) \rangle &= \lambda_1 \\
\langle d\phi_{p}(e_1), d\phi_{p}(e_2) \rangle &= \mu \\
\langle d\phi_{p}(e_2), d\phi_{p}(e_2) \rangle &= \lambda_2
\end{align*}
Now take:
\begin{align*}
v_1 &= e_1 \\
v_2 &= \cos\theta\ e_1 + \sin\theta\ e_2
\end{align*}
The equation in your question implies that:
$$
\cos\theta = \frac{\lambda_1 \cos\theta + \mu \sin\theta}{\sqrt{\lambda_1\left(\lambda_1\cos^2\theta + 2\mu\sin\theta\cos\theta + \lambda_2\sin^2\theta\right)}}
$$
Take $\theta = \frac{\pi}{2}$ to get $\mu = 0$. This implies that:
$$
\lambda_1 = \lambda_1 \cos^2\theta + \lambda_2\sin^2\theta
$$
Or $\lambda_1 = \lambda_2$. Hence:
\begin{align*}
\langle d\phi_{p}(e_1), d\phi_{p}(e_1) \rangle &= \lambda_1 \langle e_1, e_1 \rangle_{p} \\
\langle d\phi_{p}(e_2), d\phi_{p}(e_2) \rangle &= \lambda_1 \langle e_2, e_2 \rangle_{p} \\
\langle d\phi_{p}(e_1), d\phi_{p}(e_2) \rangle &= \lambda_1 \langle e_1, e_2 \rangle_{p} \qquad (= 0)
\end{align*}
Since both $\langle, \rangle_{p}$ and $\langle d\phi_{p}(), d\phi_{p}() \rangle$ are bilinear forms, the above is true for all $v_1, v_2 \in T_{p}S$.
Theorem. Let $C(M)$ be the conformal group of a Riemannian manifold $M$ with $dim(M)=n \ge 2$. If $M$ is not conformally equivalent to $S^n$ or $E^n$, then $C(M)$ is inessential, i.e. can be reduced to a group of isometries by a conformal change of metric.
This theorem has a long and complicated history, you can find its proof and the historic discussion in
J. Ferrand, The action of conformal transformations on a Riemannian manifold. Math. Ann. 304 (1996), no. 2, 277–291.
In view of this theorem, whenever $f: (M,g)\to (M,g)$ is a conformal automorphism without a fixed point, then there exists a positive function $\alpha$ on $M$ such that $f: (M,\alpha g)\to (M,\alpha g)$ is an isometry. I will prove it in the case when $(M,g)$ is conformal to the sphere and leave you the case of $E^n$ as it is similar.
Every conformal automorphism $f$ of the standard sphere which does not have a fixed point in $S^n$ has to have a fixed point $p$ in the unit ball $B^{n+1}$. (I am using the Poincare extension of conformal transformations of $S^n$ to the hyperbolic $n+1$-space in its unit ball model.) After conjugating $f$ via an automorphism $q$ of $S^n$ (sending $p$ to the center of the ball), we obtain $h=q f q^{-1}$ fixing the origin in $B^{n+1}$, which implies that $f\in O(n+1)$ and, thus, preserves the standard spherical metric $g_0$ on $S^n$. Now, use the fact that $g_0$ is conformal to $g$ and $q^*(g_0)$ is conformal to $g$ as well. qed
Best Answer
Hint: If $\phi$ is conformal and area-preserving, and if $dA$ and $d\bar{A}$ denote the respective area forms of $S$ and $\bar{S}$ (in the metrics induced by the embeddings in $\mathbf{R}^{3}$), then for every region $R \subseteq S$, \begin{align*} \iint_{R} \phi^{*} d\bar{A} & = \iint_{\phi(R)} d\bar{A} && \text{change of variables} \\ &= \operatorname{area}(\phi(R)) \\ &= \operatorname{area}(R) && \text{$\phi$ is area-preserving} \\ &= \iint_{R} dA. \end{align*} Use the fact that $\phi$ is conformal to express $\phi^{*} d\bar{A}$ in terms of $\lambda$ and $dA$, then use the facts that $\lambda$ is continuous and $R$ is arbitrary to consider what happens if $\lambda^{2}$ is not identically equal to $1$.