Luckily, the octonions are 'alternative': the unital subalgebra generated by any two octonions is associative. Furthermore, the inverse 𝑔⁻¹ of any nonzero octonion is in the unital subalgebra generated by 𝑔. Thus 𝑔, 𝑥 and 𝑔⁻¹ all lie in a unital subalgebra generated by two octonions, so

(𝑔𝑥)𝑔⁻¹ = 𝑔(𝑥𝑔⁻¹)

and we can write either one as 𝑔𝑥𝑔⁻¹ without fear.

So the big question is whether

𝑓(𝑥) = 𝑔𝑥𝑔⁻¹

is an automorphism - that is, whether it obeys 𝑓(𝑥𝑦)=𝑓(𝑥)𝑓(𝑦).

(2/n)

In other words: if we have a nonzero octonion 𝑔, do we have

𝑔(𝑥𝑦)𝑔⁻¹ = (𝑔𝑥𝑔⁻¹) (𝑔𝑦𝑔⁻¹)

for all octonions 𝑥 and 𝑦? Again this is not obvious, because the octonions are nonassociative!

And indeed it's not always true.

This paper:

P. J. C. Lamont, Arithmetics in Cayley's algebra, Glasgow Mathematical Journal, 6 no. 2 (1963), 99-106. https://www.cambridge.org/core/services/aop-cambridge-core/content/view/F911E6DABC9FCABB17A531156AD272AF/S2040618500034808a.pdf/div-class-title-arithmetics-in-cayley-s-algebra-div.pdf

claims to settle when it's true! And the answer given is interesting.

(3/n)

Namely, a nonzero octonion 𝑔 has

𝑔(𝑥𝑦)𝑔⁻¹ = (𝑔𝑥𝑔⁻¹) (𝑔𝑦𝑔⁻¹)

for all octonions 𝑥 and 𝑦 precisely when 𝑔 lies at a 0 degree, 60 degree, 120 degree or 180 degree angle from the positive real line. (We think of the real multiples of 1 ∈ 𝕆 as a copy of the real line in the octonions.)

The proof is a terse calculation - but I haven't checked it carefully yet.

(4/n)

In particular, 𝑔 is a 6th root of 1 in the octonions if and only if |𝑔| = 1 and 𝑔 lies at a 0, 60, 120 or 180 degree angle from the positive real line.

So, if Lamont is correct, 6th roots of 1 give inner automorphisms of the octonions! In fact they give all the inner automorphisms, since rescaling 𝑔 goesn't change 𝑔𝑥𝑔⁻¹.

1 and -1 are 6th roots of 1 that give trivial inner automorphisms - just the identity. But the rest are nontrivial!

(5/n)

@undefined @johncarlosbaez they generate the whole G_2. Lie subgroups of G_2 are well understood. as the set of inner automorphisms you describe is 6-dimensional the only possibilities we need to exclude is that it generates a copy of SU(3) which is 8 dimensional or a copy of SO(4) (all other proper subgroups of G_2 have dimensions <6). It should be easy to see that it's not SO(4) which is also 6-dimensional.

SU(3) can be excluded as follows.
All SU(3) in G_2 are known to come about as follows. G_2 acts on S^6=unit imaginary octonians and various SU(3) in G_2 are isotropy subgroups of different points in S^6. if all inner automorphisms were in an isotropy group of some point v in S^6 that v would be in the center of octonions. This is impossible and hence Inner automorphisms generate all of G_2.