Given a manifold $M$, and points ${\bf p}, \ {\bf q} \in M$, the logarithm of ${\bf p}$ with respect to ${\bf q}$, $ \ \textrm{log}_{\bf q} ({\bf p})$, is an element of $T_{\bf q}M$ giving the position of the point ${\bf p}$ relative to ${\bf q}$, i.e. the inverse of the exponential map.
Let $\phi: M \rightarrow M'$ be a diffeomorphism between $M$ and another manifold $M'$. From my understanding, if $\phi$ is an isometry, then $$\left[d \phi\right]\vert_{\bf q} \cdot \textrm{log}_{\bf q} ({\bf p}) = \textrm{log}_{\phi({\bf q})}\left( \phi({\bf p}) \right) \in T_{\phi({\bf q})}M',$$ where $\left[d \phi \right]\vert_{\bf q}$ denotes the differential of $\phi$ evaluated at ${\bf q}$. Intuitively, this makes sense since isometries preserve distances and angles and $\left[d \phi\right]$ is an orthogonal transformation.
Does the above equality hold when $\phi$ is a conformal transformation? My hunch is that it doesn't, since I suspect the geodesic distance between $\phi({\bf q})$ and $\phi({\bf p})$ is a non-linear function due to the non-constant scale factor.
If not, how does the logarithm map transform when $\phi$ is a conformal transformation?
EDIT
After some thought, I think my question boils down to something simpler:
Do conformal transformations preserve direction of the the logarithm map, i.e. the angle of the position of ${\bf p}$ relative to ${\bf q}$ in geodesic polar coordinates?
At first glance, this seems like it might be true, since conformal mappings preserve angles in the tangent space, though Anthony's comment below could for the basis of a possible counter-example.
If true, then the logarithm could transform via the following equality: $$ \textrm{log}_{\phi({\bf q})}\left(\phi({\bf p})\right) = \frac{\lvert [ d \phi ] \vert_{\bf q} \rvert^{-\frac{1}{2}} \cdot g_{M'} \left(\phi({\bf p}), \ \phi({\bf q}) \right) }{g_M \left({\bf p}, \ {\bf q} \right)} \cdot \left[d \phi \right]\vert_{\bf q} \cdot \textrm{log}_{\bf q} ({\bf p}),$$ where $g_{M}$ and $g_{M'}$ denote the geodesic distances on $M$ and $M'$, respectively, and $\lvert \ \cdot \ \rvert$ is the determinant.