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Sep 11, 2017 at 12:56 comment added Peter Heinig You meant to ask for the isomorphism type of $U(\mathrm{Aut}(\Gamma))$, described in usual group theoretic discourse. In particular, by the formula for the cardinality of finite conjugacy classes, $\mathfrak{C}$ has cardinality $\frac{6!}{2^2\cdot 2! \ \cdot \ 1^2\cdot 2!} = 45$, so the $U(\mathrm{Aut}(\Gamma)$ described above is a subgroup of $\mathrm{Sym}(\text{$45$-element finite set})$, in other words, is a permutation group on 45 letters. In other words, your graph has 45 vertices. You probably essentially know this. I was just trying to correct my muddled comments of yesterday.
Sep 10, 2017 at 10:21 comment added Peter Heinig An article worth pointing out in this context, despite the incorrect grammar in the article's title, seems to be Mahsa Mirzargar, Peter P. Pach, and A. R. Ashrafi: The automorphism group of commuting graph of a finite group. Bull. Korean Math. Soc. 51 (2014), No. 4, pp. 1145–1153. This article, however, seems not to contain a direct answer to OP: all results in loc. cit. are about commuting graphs on a group in its entirety, not on a conjugacy class, as in the OP. In particular, YCor's answer presently is the only answer known to me.
S Sep 10, 2017 at 10:11 history suggested Peter Heinig CC BY-SA 3.0
Various stylistic and grammatical improvements. Style of question was not preserved, yet it seems that the errors were too many, and in the new form this question is more useful to others. (In particular, with the keyword "commuting graph" in the new title.) Content preserved.
Sep 10, 2017 at 9:29 review Suggested edits
S Sep 10, 2017 at 10:11
Sep 10, 2017 at 7:48 answer added YCor timeline score: 9
Sep 10, 2017 at 7:42 answer added F. C. timeline score: 6
Sep 10, 2017 at 7:06 review First posts
Sep 10, 2017 at 7:37
Sep 10, 2017 at 7:01 history asked maryam CC BY-SA 3.0