Q:

Time dilation and symmetry in special relativity

Trying to grasp special relativity concepts, I thought in the following experiment.
Imagine Alice took a trip in a spaceship to another star. Now, she is returning close to light speed. When she passes by Pluto, she is no longer accelerating. In this exact moment, she starts her chronometer at 0:00:00.
Meanwhile, on Earth, Bob has a telescope pointing to Pluto. The moment he sees Alice's spaceship passing Pluto, he also starts his chronometer, but instead of starting at 0:00:00, he compensates for the time light took to travel from Pluto to Earth.
Alice isn't aiming exactly at Earth, so she is not slowing down. When she passes by Earth, she takes a Picture from Bob's chronometer, while Bob takes a picture from her chronometer.
Question: which chronometer will be late? The two observers should perceive each other's time running slow. Which one is correct?
Considerations:

Suppose Earth and Pluto are aligned in relation to Alice path.
Earth's and Sun's gravitational effects should be ignored to avoid general relativity issues. The same applies to Earth orbital translation.

A:

Here is you basic problem

The moment [Bob] sees Alice's spaceship passing Pluto, he also starts his chronometer, but instead of starting at 0:00:00, he compensates for the time light took to travel from Pluto to Earth.

The compensatory offset, of course, depends on the distance from Pluto to Earth but Alice and Bob don't agree on that figure (length contraction, right?). So, to Alice, Bob has offset his clock by too much, but his clock is also running slow by a similar fraction.
The result is that they will agree on what their images should show (which is easier to work out in Bob's frame of reference, I think).
As usual things are easier in terms of invariants. Both Alice and Bob can compute the interval between Alic's passing the two planets 
$$(\Delta s)^2 = (\Delta t)^2 - (\Delta x)^2 \,,$$
and get the same value. And because $(\Delta x)_A \equiv 0$, that implies that
$$ (\Delta t)_A^2  < (\Delta t)_B^2 \,.$$
Both parties compute that using this procedure, Bob's clock will be ahead of Alice's when Alice passes Earth.
Mind you, this doesn't break the symmetry of time dilation. 
Here's why: if instead of working the problem in terms of Alice's ship passing planets, Alice deploys a buoy in front of her ship at a distance so that (in her frame) it passes Earth as she passes Pluto and both participants take their their start time from the buoy passing Earth the problem is reversed. To emphasize the change: in the original version of the problem both events involved Alice's ship, but Earth and Pluto were involved in only one event each; in the new version, both event involve Earth, but the buoy and Alice's ship are only involved in one event each. In the original the frame of Alice's ship (involved in both events) experiences the least time; in the modified version the frame of Earth (involved in both events) experiences the least time.
What is happening in both these examples is that you are rigging the pair of events timed to be spatially coincident in one of the frames which makes that frame special for that particular measurement (but not special in general). In general it is not safe to try to reason about time-dilation or length-contraction in isolation from one another and is generally easier and safer to reason about intervals instead.

