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Feb 12, 2021 at 2:15 comment added Richard Lyons $B_3=Z\times S$ where $|Z|=2$ (generated by the -1 mapping) and $S=D_3\cong S_4$, the symmetric group. There is a unique surjective homomorphism $\sigma:S\to Z$, the sign homomorphism. Then the mapping $\alpha:(z,s)\mapsto (z\sigma(s),s)$ is an automorphism of $Z\times S\cong B_3$. There are reflections in $S$, and they are carried by $\alpha$ to their negatives, so $\alpha$ is not inner -- it's not even conjugation by any element of $GL_3(R)$.
Feb 12, 2021 at 1:06 comment added M. Winter @RichardLyons Can you say more about this?
Feb 12, 2021 at 0:50 comment added Will Sawin @RichardLyons Good point, thanks.
Feb 12, 2021 at 0:50 history edited Will Sawin CC BY-SA 4.0
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Feb 12, 2021 at 0:41 comment added Richard Lyons I believe B_3 does have an outer automorphism, but not in $O(R^3)$.
Feb 11, 2021 at 16:17 history answered Will Sawin CC BY-SA 4.0