= Solution
The <conformal invariance of planar Brownian motion> is a time-change statement. If $\phi:D\to D'$ is a conformal bijection of planar domains and $B$ is <planar Brownian motion> started in $D$, let $\tau_D$ be its exit time and define
$$
A_t=\int_0^{t\wedge\tau_D}|\phi'(B_s)|^2\,ds.
$$
Before exit this clock is strictly increasing. Its inverse $C_u$ makes $\phi(B_{C_u})$ a <planar Brownian motion> started at $\phi(B_0)$, up to its transformed lifetime $A_{\tau_D}$. The image of the stopped trajectory is therefore invariant in law after this change of time. A holomorphic map with nonzero derivative gives the corresponding local statement even without global injectivity.
Here is the proof. Write $\phi=u+iv$ and $B=(B^1,B^2)$. The real and imaginary parts are harmonic. The <Itô formula> therefore gives the two <continuous local martingales>
$$
u(B_t)-u(B_0)=\int_0^t\nabla u(B_s)\cdot dB_s,
\qquad
v(B_t)-v(B_0)=\int_0^t\nabla v(B_s)\cdot dB_s,
$$
stopped on compact subsets of $D$. The <Cauchy-Riemann equations> imply
$$
|\nabla u|^2=|\nabla v|^2=|\phi'|^2,
\qquad\nabla u\cdot\nabla v=0.
$$
Thus both coordinate <quadratic variations> equal $A_t$ and their <quadratic covariation> is zero. After inverse time change, the covariance matrix of the two <continuous local martingales> is $uI_2$. The multidimensional <Lévy characterization of multidimensional Brownian motion> states that a continuous vector <local martingale> starting at zero with this covariance matrix is standard vector <Brownian motion>. Apply it after subtracting $\phi(B_0)$, and exhaust $D$ by compact subsets. This proves the theorem, with killing at the transformed exit lifetime. The clock is the <conformal Brownian clock>; omitting it would generally give an incorrect speed.
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