@guohua As long as min not equals to max, you keep doing add 1 to (n-1) elements, then min value for sure +1, I believe you understand this part already. So, I suppose your question is what if m moves means min value once equals to max and somehow this array adds 1 to (n-1) elements which makes it to go for another round to have all equal elements. If you think carefully, you will find out that by doing this, there exists a single add-1-operation that min value doesn't get to +1.

So, if you want to get the second minimum moves to reach to all equal state, then you can have these equations instead:(I am using OP's equations so I think you can better understand and compare the differences)

for second minimum m moves:

sum+m*(n-1)=x * n

x=minNum+m-1 (there is a single time min value doesn't +1)

sum-minNum * n+n=m (m gets n moves more compared to the minimum value, because it also means you need to decrease every element by 1 after it reaches to the first equality, easy to understand, right?)