On the automorphism group of a binary self-dual $$[120, 60, 24]$$[120,60,24] code

We prove that an automorphism of order 3 of a putative binary self-dual $$[120, 60, 24]$$[120,60,24] code $$C$$C has no fixed points. Moreover, the order of the automorphism group of $$C$$C divides $$2^a\cdot 3 \cdot 5\cdot 7\cdot 19\cdot 23\cdot 29$$2a·3·5·7·19·23·29 with $$a\in \mathbb N _0$$a∈N0. Automorphisms of odd composite order $$r$$r may occur only for $$r=15, 57$$r=15,57 or $$r=115$$r=115 with corresponding cycle structures $$3 \cdot 5$$3·5-$$(0,0,8;0), 3\cdot 19$$(0,0,8;0),3·19-$$(2,0,2;0)$$(2,0,2;0) or $$5 \cdot 23$$5·23-$$(1,0,1;0)$$(1,0,1;0) respectively. In case that all involutions act fixed point freely we have $$|\mathrm{Aut}(C)| \le 920$$|Aut(C)|≤920, and $$\mathrm{Aut}(C)$$Aut(C) is solvable if it contains an element of prime order $$p \ge 7$$p≥7. Moreover, the alternating group $$\mathrm{A}_5$$A5 is the only non-abelian composition factor which may occur in $$\mathrm{Aut}(C)$$Aut(C).