Conserved?

When Linear Momentum is conserved then, $p=constant$ or $dp/dt =0$ , hence $F_{ext}$ =0.

Now $F_{ext}$ =0 then Whatever be the axis of rotation , $T_{ext}$ =0. or $dJ /dt = 0$ or $J = constant$

Let $\vec{p}(t)$ and $\vec{L}(t)$ respectively be the linear and angular momentums of the particle of mass $m$ . Given, $\frac{d\vec{p}(t)}{dt}=\vec{0}$ Now we know that $\vec{L}(t)= \vec{r(t)} \times \vec{p}(t)$ , where $\vec{r}(t)$ is the position-vector of the particle with respect to any fixed point and $'\times'$ denotes the vector cross-product operation. Then, using product rule for vector differentiation we have: $\frac{d\vec{L}(t)}{dt}= \vec{r}(t) \times \frac{d\vec{p}(t)}{dt}+ \frac{d\vec{r}(t)}{dt} \times \vec{p}(t)=\vec{r}(t) \times\vec{0}+\frac{1}{m} \vec{p}(t) \times \vec{p}(t) \stackrel{(a)}{=} \vec{0},$ where in (a) we have used the fact that the cross product of two parallel vectors is zero.