Geometric reason why conjugation by an element in $B_3$ inverts this element?

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Let $B_3$ be the braid group on three strands. I was looking at an element in $B_3$, which I will write in the standard presentation:

$$(\sigma_2\sigma_1\sigma_2)^{-1}\sigma_1^3\sigma_2^{-3}(\sigma_2\sigma_1\sigma_2)$$

and I was able to explicitly show it equal to $\sigma_2^{3}\sigma_1^{-3}$, which inverts the element by conjugation. I was wondering if one see this geometrically? (Via some diagram) Or rather, if there is some phenomenon that explains this, or if it is a mere coincedence.

If the following is known: what do inner automorphisms of $B_3$ look like in general? Maybe the semi direct product presentation is more promising for understanding it.

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This braid group is a central extension of the modular group $PSL(2,Z)$. In the case of the latter, it is easy to see where this phenomenon comes from: An order two rotation of the hyperbolic plane which preserves the axis of a hyperbolic element and swaps its endpoints. To see how this can happen, lift to $SL(2,Z)$. The order two rotation lifts to an order 4 rotation. It will conjugate a real-diagonalizable matrix $M\in SL(2,Z)$ to its inverse iff the eigenlines of $M$ are orthogonal, i.e. iff $M=M^T$. There are infinitely many examples of this happening.

As for your last question, for every group $G$, $Inn(G)\cong G/Z(G)$, where $Z(G)$ is the center of $G$. In the case of $B_3$, this quotient is isomorphic to $PSL(2,Z)$.

See this Wikipedia article for more detail.