Let Aut$ (M)$ be the automorphism group ($M$ isomorphisms in and of itself). Show That there are natural numbers $n_1,. . . , n_k$ such that the number of elements of Aut $(M)$ is $n_1! · · · n_k!$.
The ideal solution to this problem is to understand it too well, but I fail to grasp the essence of what they ask to try, could someone explain to me the way forward and give me an explanation of what this result means, please?