It depends on what you know about the automorphism group of a cyclic group.

1. Aut( ) is isomorphic to the group of (multiplicative) units of .

2. If is the factorization of n into primes, Aut( ) is isomorphic to

3. If p is odd is isomorphic to

4. A similar statement for n a power of 2 (I'm tired of typing Latex).

All of the above is easily found on the web -- use Google

I'll assume just 1. So we need the group multiplicative units of ; i.e. the set of integers < 24 which are relatively prime to 24, namely {1,5,7,11,13,17,19,23}. Now

, etc (equality here is read mod 24). That is all elements have order 2. Now you must figure that out that such a group is the direct sum of 3 cyclic groups of order 2. One "high powered" result: If an abelian group G has order p^n for a prime p and every element has order p, G is isomorphic to the direct sum of n cyclic groups of order p. This is one part of the fundamental theorem of finite abelian groups.

I really think the general proofs of the statements above are no harder than trying to do the specific n = 24.