The issue of tachyonness is a red herring; and the question is entirely a matter of simple computation, that you should do for yourself.
Here it is.
Let $x = v t$ describe the motion of an object in a direction parallel to the $x$ axis, where $t$ is the time, where the origin on the $x$ axis is where you are at and where time $t = 0$ is when the object is where you are at. It is moving with a speed $v$, which is negative if it is toward you, positive if it is moving away from you.
We'll also assume you're staring in the direction of increasing $x$. A signal moving at the light speed $c$ has a motion given by $x = b - c t$, if it is coming at you from the direction of increasing $x$. If it is at the location of the object at $(x, t) = \left(x_0, t_0\right)$ and at your location at $(x, t) = \left(x_1, t_1\right)$, then
$$
x_0 = b - c t_0, \hspace 1em x_0 = v t_0,\\
x_1 = b - c t_1, \hspace 1em x_1 = 0.
$$
Therefore $(v + c) t_0 = b = c t_1$, and:
$$(v + c) x_0 = (v + c) v t_0 = v (v + c) t_0 = v c t_1.$$
The motion you see is given by the stream of signals and is therefore described by $(x,t) = \left(x_0, t_1\right)$. You could do this extra carefully by trying to separately account for left and right eyes and use triangulation to determine the "apparent" distance in place of $x_0$, but I don't think that extra complication will significantly affect the issue (maybe except when it's close), so we'll just use $x_0$, itself.
Thus, assuming $v ≠ -c$, and $v ≠ 0$, we have
$$x_0 = \frac{v c}{v + c} t_1 = \frac{c}{1 + c/v} t_1 = \frac{v}{1 + v/c} t_1,$$
and you see the object moving with a speed $c/(1 + c/v) = v' = v/(1 + v/c)$.
If $v = 0$, then $x_0 = v t_0 = 0$ and you see the object up right next to you, so $v' = 0$.
If $v = -c$, then $b = 0$ and, thus, $t_1 = 0$, and that's the only signal you'll ever see, and you'll only see it in a flash at time 0, and $v' = ∞$.
If $v > 0$, then $0 < v' < c$. If $-c < v < 0$ then $v' < 0$. If $v < -c$ then $v' > 0$. If $v = ∞$ then $v' = c$. An object going at the speed $v = ∞$ will appear to you to be moving away from you at light speed - even if you turn around and look at it in the other direction.
By the way, notice that I actually said nothing about Relativity in the determination of $v'$. Relativity is also a red herring to the question; it's the same answer as in non-relativistic physics. You get the same expression for $v'$, regardless of which paradigm you are in, it is paradigm-independent, and the question actually has nothing to do with Relativity, per se, but merely with signal processing. You could just as well take any other speed $V$ in place of $c$ (e.g. the speed of light in water) to determine the visual effect of an object's motion, in which case you'll get $v/(1 + v/V) = v' = V/(1 + V/v)$ in place the expression previously derived - again, independent of paradigm. So, even light speed is a red herring here.
