We know that the four-dimensional scalar product is invariant under coordinate transformation, hence the space-time interval and proper time is also invariant. Since the 4-velocity is given by space-time interval divided by proper time. Then why 4-velocity is not invariant?
Please help me to clarify my conceptual misunderstanding. Any help is appreciable.
As you say individual coordinate positions $\vec{s}$ are not invariant, only the magnitude of spacetime intervals $\Delta \vec{s}\cdot\Delta\vec{s}$ are invariant.
Similarly, the coordinate velocity $\vec{u}$ is not invariant. Different observers moving at different speeds with their axes oriented in different directions will not be able to agree on the individual components of the coordinate velocity of an object. They will all agree on the magnitude of the 4-velocity, that can be found from the scalar product $\vec{u}\cdot\vec{u}$.
4-velocity is defined $\frac{d\vec{s}}{d\tau}$. Remember, $d\vec{s}$ alone is not invariant, but $d\vec{s}\cdot d\vec{s}$ is. $|\vec{u}|^2 = \frac{d\vec{s}}{d\tau}\cdot \frac{d\vec{s}}{d\tau}$ is too.
We know that the four-dimensional scalar product is invariant under coordinate transformation, hence the space-time interval and proper time is also invariant.
Correct. Space-time interval in $(-,+,+,+)$ signature is $ds^2 = -c^2dt^2+dx^2+dy^2+dz^2$ and is Lorentz invariant, and by definition, proper time $d\tau = \frac{\sqrt{-ds^2}}{c}$ is also invariant.
Since the 4-velocity is given by space-time interval divided by proper time. Then why 4-velocity is not invariant?
No, 4-velocity is $\vec u = \frac{\vec {ds}}{d\tau} = (\gamma, \gamma \frac{\vec v}{c})c$. Now this cannot be invariant for all cases, precisely because of time-dilation. It can be in case of time-dilation factor being one or equivalently $v = 0$ where the observer's frame of reference is not moving at all. This should be true, just as expected.
