In his classic paper on Hecke algebra representations of braid groups from 1987, Vaughan Jones makes the claim that "the various generators $\sigma_i$ are all conjugate." How does one see this?
I think an example showing that $\sigma_1$ and $\sigma_2$ are conjugate would be convincing enough.
On
Ah, $\sigma_1$ and $\sigma_2$ is easy. In the braid group $B_{n+1}$ on $n+1$ strands, take $\sigma_1 \cdots \sigma_n$. We'll show that $$(\sigma_1 \cdots \sigma_n)\sigma_1(\sigma_1 \cdots \sigma_n)^{-1} = \sigma_2$$ By repeated application of far commutativity, we have \begin{align*} \sigma_1 \cdots \sigma_n \sigma_1 \sigma_n^{-1} \cdots \sigma_1^{-1} &= \sigma_1\sigma_2\sigma_1\sigma_3\cdots\sigma_n\sigma_n^{-1} \cdots \sigma_1^{-1} \\ &= \sigma_1\sigma_2\sigma_1\sigma_2^{-1}\sigma_1^{-1}. \end{align*}
Applying the braid relation $\sigma_i\sigma_{i+1}\sigma_i = \sigma_{i+1}\sigma_i\sigma_{i+1}$ gives the desired result.
Now that I've seen this, I can generalize a bit: conjugation of $\sigma_i$ by $\sigma_i \cdots \sigma_n$ (for $i < n$) gives $\sigma_{i+1}$. Actually, in fact, by the same argument as above, except with applications of far commutativity again at the very end, one can see that
$$(\sigma_1 \cdots \sigma_n)\sigma_i(\sigma_1 \cdots \sigma_n)^{-1} = \sigma_{i+1}$$
For adjacent generators, an easier way to see that they are conjugate is to use $\sigma_i\sigma_{i+1}.$ Then after only applying the braid relation, we'll get $$(\sigma_i\sigma_{i+1})\sigma_i(\sigma_i\sigma_{i+1})^{-1} = \sigma_{i+1}.$$
EDIT: Another thought
Conjugation of $\sigma_1$ by the garside braid gives $\sigma_n.$
Here is a conjugate of $\sigma_2$ that is equal to $\sigma_1$:
Algebraically, this is the relation $\sigma_2\sigma_1\sigma_2\sigma_1^{-1}\sigma_2^{-1} = \sigma_1$, which is just another way of writing the braid relation. A similar argument shows that $\sigma_i$ and $\sigma_{i+1}$ are conjugate for each $i$, which proves that all of the generators are conjugate to one another.