Thank you. Got it.

In[1]:= LCM[10, 12]

Out[1]= 60

The determinant will have to be an nth root of unity. So I'd try powering to multiples of n. Of course this might just be 1, in which case it provides no shortcut. Working with the eigenvalues might be faster than brute-force powering, I'm not sure. By this I mean deducing the orbit length as LCM of the roots of unity of the diagonalized form (with some abuse of notation in that remark).

Without loss of generality, we can work with 2-dimensional matrices.

The Abelian group of order 3 has these elements:

In[429]:= A3 = {IdentityMatrix[2], RotationMatrix[2 \[Pi]/3], 
  RotationMatrix[4 \[Pi]/3]}

Out[429]= {{{1, 0}, {0, 1}}, {{-(1/2), -(Sqrt[3]/2)}, {Sqrt[3]/
   2, -(1/2)}}, {{-(1/2), Sqrt[3]/2}, {-(Sqrt[3]/2), -(1/2)}}}

We can define a function to compute the order of a group element by iterating over matrix multiplication:

In[430]:= groupElementOrder[m_?MatrixQ] := 
 Module[{n = 1, e = IdentityMatrix[Length[m]]}, 
  NestWhile[(n++; Dot[#, m]) &, m, # != e &]; n]

NestWhile does the iteration, and terminates when the matrix product equals the identity matrix. Now, we can test it on the group A3:

In[431]:= groupElementOrder /@ A3

Out[431]= {1, 3, 3}

I am not sure exactly what you are asking, but perhaps this might give you an idea for a direction.

Here is the finite list generated by repeated multiplications by I:

FixedPoint[Union[Join[#, I #]] &, {1}, 1000]

So the order is:

Length[  FixedPoint[Union[Join[#, I #]] &, {1}, 1000]]

For multiplications by Exp[Pi I/12]:

FixedPoint[Union[Join[#, Exp[Pi I/12] #]] &, {1}, 10000]