Would a navigator announcing ship velocity whle approaching lightspeed make linear announcements? Given my admittedly limited understanding of relativity, I believe that as a hypothetical space ship approaches the speed of light at a constant acceleration, what the crew would "see" on Earth (if they could) is life speeding up until it's well beyond a blur.  What the people of Earth would see if they could "see" into the ship are people slowing down until they stop moving.
My question is this:

*

*Given a completely unrealistic constant acceleration of 0.1c/60s

If the navigator announces the passage of each 0.1c mark (0.1c... 0.2c...), will that announcement be:1

*

*Ten equally-spaced announcements (e.g. one announcement every 60 seconds)?

*Ten linearly-shorter spaced announcements (e.g., the announcements come in ever shorter-spaced increments such as Y=nX)?

*Ten exponentially-shorter spaced announcements (e.g., the announcements come in substantially shorter-based increments such as Y=nX2+mX)?

Where the functions involving variables Y and X are loosely and liberally relating the axis of velocity (Y) to the axis of time (X).
My question was born after reading this article, which suggests that from the perspective of a photon's reference frame, it is absorbed instantly after its emitted. In other words, it doesn't experience the passage of time regardless of distance. If this is in my head correctly (and based on my perspective from my initial paragraph), as the hypothetical ship increases its velocity, the experience of time between two velocities must get shorter.
I therefore think the answer is not #1.
Clarifications:

*

*My question is whether or not the navigator's perception of the increasing speed becomes non-linear as the velocity of the ship approaches light speed.


*Issues like the amount of energy needed to approach light speed, whether or not mass can hit light speed, etc, are irrelevant to my question. Horses aren't spherical, either, but my college physics professors 30 years ago talked about them a lot.

1 I realize that the announcement of "lightspeed!" will only be heard after the ship begins to decelerate from 1.0c. I'm simplifying for the purpose of the question.
 A: 
what the crew would "see" on Earth (if they could) is life speeding up until it's well beyond a blur. What the people of Earth would see if they could "see" into the ship are people slowing down until they stop moving.

No: if the ship is approaching Earth, then people on Earth will see clocks on the ship running fast, and people on the ship will see clocks on Earth running fast; while if the ship is receding from Earth, then both groups will see the other's clocks running slow. The speed ratio is not given by the time dilation factor ($γ$), but by the Doppler shift factor ($1{+}z$).
If the ship circles Earth at a constant distance, then ship clocks will run slow as seen from Earth and Earth clocks will run fast as seen from the ship. In this case the ratio is given by $γ$.

Given a constant acceleration of 0.1c/60s, if the navigator announces the passage of each 0.1c mark (0.1c... 0.2c...), will that announcement be:

The usual definition of "constant acceleration" in special relativity is constant proper acceleration, which means that those on the ship would feel a constant effective gravitational force. If the acceleration is constant in that sense, then the time between announcements will grow longer, not shorter. It will grow quadratically at first, but the time between the 0.9c and 1.0c announcements is infinite: the ship will never reach the speed of light.
You may have been thinking of a quasi-Newtonian kind of constant acceleration, where the ship's position with respect to some inertial reference frame is $x(t)=\frac12at^2$. In that case, the time between announcements will decrease (quadratically at first), but there still won't be a speed-of-light announcement because the ship can't reach the speed of light. $\frac12at^2$ acceleration simply isn't sustainable; it corresponds to a proper acceleration that goes to infinity in a finite time.
A: As the navigator and captain are in the same reference frame, there will be no relativistic time dilation between the two.
However, to accelerate an object to the speed of light would require an infinite amount of energy, and the energy required to keep accelerating increases with speed, so a constant acceleration of $0.1c$/min is not sustainable.
A: The reference frame of the ship is non-inertial, so the positioning of the navigator and the captain with respect to the acceleration direction matters.
By the equivalence principle, the ship's reference frame will look the same as being on the surface of a planet with a very strong surface gravity (51 thousand $g$s with the acceleration you have given!).
So, there will be time dilation! This, however, is only as long as there is spacing between the navigator and the captain along the acceleration axis. If this distance is $h$ then the time-dilation can be expanded in powers of $ah / c^2$, where $a$ is the acceleration. As long as $h$ is of the order of single meters this is very small so only the first term matters, so there will be dilation expressed by the formula:
$$
\Delta t_N = (1 - ah/c^2) \Delta t_C
\,.
$$
If the navigator is "above" the captain, it will appear "blueshifted": the impulses it sends will seem closer to each other to the captain.
This would be your option 2: note, however, that if the captain is "above" the navigator the effect works in the opposite direction, and the pulses will be more spaced.
Anyway, the formula is always linear.
All of this applies from the beginning, there is no need for the speed to approach $c$; as other answers clarified accelerations do not work as in Newtonian mechanics for relativistic speeds.
This idea of acceleration, however, works well enough if we interpret it as a Taylor expansion near $v/c=0$.
Since speeds are relative, the effect will be the same as long as the engines of the spaceship keep applying the same thrust and as long as the navigator sends impulses which are equally spaced in time.
As the speed increases, even with constant thrust the multiples of $0.1c$ will not be reached linearly.
A reasonable form for the position as a function of time of an object with constant acceleration is
$$
x(t) = \frac{c^2}{a} \left(\sqrt{1 + \frac{a^2 t^2}{c^2}} - 1\right)
$$
so the velocity looks like
$$
v(t) = \frac{at}{\sqrt{1 + a^2 t^2 / c^2}}
$$
which, as you can see, approaches $c$ asymptotically for large $t$, but never reaches it.
For small $t$, it is a good approximation to say $v = at$.
Edit:
Since you're asking about perception, let us explore this a bit further, starting from the expression I wrote for the velocity.

Here is a plot of it, you can see that even though the acceleration is constant the speed approaches $c$ asymptotically.
However, there is an important effect we must consider, which is the reason why this might seem weird: length contraction.
An observer moving at a certain velocity will see objects contracted along the direction of motion by a factor
$$
\gamma = \frac{1}{\sqrt{1 - v^2 / c^2}}
\,.
$$
So, even though the speed approaches $c$, the distance elapsed per second in the reference frame of the Earth (or anyhow, the frame in which the rocket started) can be larger than $c$.
Here is a plot of how $\gamma$ looks, for the same times as the velocity plot:

You can see that as the speed approaches $c$ this becomes quite large.
When we are to compute the distance elapsed in the Earth frame, we must use a sort of "effective velocity", computed as $\gamma v$: this can be much larger than $c$.
The velocity cannot increase further than $c$, so length contraction "picks up the slack": lengths are contracted ever further, so from the perspective of the rocket things whiz by at almost $c$, but they are much shorter so the speed referred to the Earth-frame length would be $>c$.
This effective velocity $\gamma v$ does indeed increase linearly with $\gamma v  = at$, as one might expect.
This is all done under the assumption that $a$ is constant, and here $a$ is a proper acceleration: the thrust of the engine, if you will.
Thus, an observer in the ship would perceive uniform acceleration.
The "photon frame" does not exist, but what could exist is the frame of an observer moving with very large $\gamma v \gg c$: they would see stationary objects as very contracted along the axis of motion, and moving very slowly.
Such an observer would be able to travel between stars (separated by $d$) in a very short subjective time ($d / \gamma v$), while the time they take as measured by observers static with respect to the stars would be bounded by $d /c$.
A: If the ship programs the engines to have an increment of 0.1c every 60s, the interval between the announcements (until 0.9 c) is by definition 60s in the ship's time . The way the ship knows its velocity can be the average velocity of stars (blue shifting ahead and red shifting behind).
Of course the crew feels an increasingly strong fictitious force to match that constant rate of speed increase. Also any loose object would fall at increased acceleration along the time. That means: in order to have an equally spaced increment of velocity, it is necessary to have an increased local acceleration.
The crew would never measure stars blue/red shifting indicating a velocity equal to c, so the last step is not possible. But with a small modification (from 0.9 to 0.999c for example) the reasoning is the same as before.
A: Mass cannot travel at c (the speed of light in a vacuum), but it could travel near c. To an outside observer considered to be at rest relative to the ship, the time passing on the ship would become more dilated (passing slower) the faster the ship went. Any one in the ships frame of reference would observe time as passing normally for them. So the people on the ship would talk normally to each other at any speed they were considered to be going, as the ship and everyone in it are in the same frame of reference.
