Is relative velocity invariant under special relativity? If a metre stick passes an observer at speed $v$, would all observers in any inertial frame of reference say the speed of the meter stick relative to the observer is exactly $v$? 
If so what is it also true for non inertial frames of reference?
 A: No, the magnitude of three-velocity is not invariant. The related quantity that is invariant is the magnitude of the four-velocity. Four-velocity is the rate of change of four-position (spacetime-position) of an object with respect to the time measured by a clock attached to the object. 
The magnitude of an object's four-velocity is always equal to $c$, the speed of light. For an object at rest (with respect to the coordinate system) its four-velocity points in the direction of the time coordinate. In contrast, the four-velocity of an object moving close to the speed of light (with respect to the coordinate system) is space-like, with a vanishing component in the time direction.
In loose terms and from a spacetime perspective, any object moves at the speed of light. Depending on the reference frame chosen this movement can be fully in the time direction (frame moving with the object), or both in time and spatial directions (any frame in which the object is not at rest).
A: Yes, there is a frame-independent concept of relative velocity. That means it's a physical concept having nothing to do with arbitrary reference frames or coordinate systems. It doesn't make sense to ask whether it's "also" invariant in noninertial frames, though you could ask whether a particular method of calculating it from frame-dependent quantities would also work in a noninertial frame.
There is a separate concept called "closing velocity" that is frame-dependent. This is simply the velocity of one of the objects minus the velocity of the other. For example, a train with speed $v$ and a beam of light heading toward each other would have a closing velocity of $v+c$ in that frame, while the ordinary relative velocity of the light relative to the train would be $c$.
A: No, and you can show this easily using the Lorentz transformations.
Start in our rest frame $S$, with an object passing the origin at velocity $u$. If the position of the object is $x = 0$ at time $t = 0$, then at time $t = T$ the position is just $x = uT$. So we have the pair of points $(0, 0)$ and $(T, uT)$.
Now look at this from the frame $S'$ moving at velocity $v$ relative to $S$, and let's calculate the velocity of the object, $u'$,  in $S'$. The Lorentz transformations tell us:
$$\begin{align}
t' &= \gamma \left( t - \frac{vx}{c^2}\right) \\
x' &= \gamma \left( x- vt \right)
\end{align}$$
If we take the origins of $S$ and $S'$ to coincide then the point $(0, 0)$ is the same in both frames, so we just need to find out where $(T, uT)$ is in S'. Substituting into the Lorentz transformations we get:
$$\begin{align}
t' &= \gamma \left( T - \frac{vuT}{c^2}\right) \\
x' &= \gamma \left( uT - vT \right)
\end{align}$$
And the velocity $u'$ is just given by $u' = x'/t'$ so:
$$\begin{align}
u' &= \frac{\gamma \left( uT - vT \right)}{\gamma \left( T - \frac{vuT}{c^2}\right)} \\
   &= \frac{u - v}{1 - \frac{vu}{c^2}}
\end{align}$$
The relative velocity of the object and $S$ is then just:
$$\begin{align}
 u'_{rel} &= u' - (-v) \\
          &= \frac{u - v}{1 - \frac{vu}{c^2}} + v \\
          &= \frac{u - v^2u/c^2}{1 - vu/c^2}
\end{align}$$
So the relative velocity of S and the object is not the same for all observers.
A: Yes, you can think of this geometrically.  In Euclidean space, the angle between two vectors is rotationally invariant.  In Minkowski space, the same applies with respect to Lorentz transformations.  Here, the "angle" is the "relative rapidity" between two different timelike vectors.
Expiclitly, we have $a \cdot b = |a| |b| \cosh \varphi$ for $\varphi$ the relative rapidity.  This inner product is a Lorentz scalar and thus invariant with respect to the Lorentz transformation.  The magnitudes $|a|$ and $|b|$ are also Lorentz invariants, so $\cosh \varphi$ must be Lorentz invariant, and therefore $\varphi$ is Lorentz invariant.  Note that $\cosh \varphi = \gamma$ and $\sinh \varphi = \gamma \beta$.
An observer following one of the timelike vectors will perceive an object following the other timelike vector as having a velocity determined by that relative rapidity.  All other inertial frames agree on what that relative rapidity is (because of LT invariance of the relative rapidity), so there is no ambiguity.
