In how many ways can we use c colors to paint cross? We have a lot of small squares, we put them into larger squares of nxn size, then use these 5 nested squares to form a cross. For example, with n = 2 we have the following image: 

In how many ways, can use c colors to paint on small squares of 1x1 size of the cross.
  Knowing that the two ways to rotate when rotating on a flat surface overlap are considered as one way.

Suppose $f(n,c)$ is the result of problem.
I'll try build recursive form, but it isn't successful.
Initially, I think $f(n,c)=c^{5n^2}$, but it may be wrong answer.
An example: With $n=1$,$c=1$ we have $f(1,1)=1$.
Another example: With $n=1$,$c=2$, we have $f(1,2)=12$. 
 A: Use the Burnside lemma with the cycle index, which for $n$ even is
$$Z(C) = \frac{1}{4}
(a_1^{5n^2} + a_2^{5n^2/2} + 2 a_4^{5n^2/4}).$$
For $n$ odd the central small square is a fixed point, giving
$$Z(C) = \frac{1}{4}
(a_1^{5n^2} + a_1 a_2^{(5n^2-1)/2} + 2 a_1 a_4^{(5n^2-1)/4}).$$
Here we have the first term  corresponding to the identity, the second
to the $180$ degree rotation and the last to the two rotations by $90$
and $270$ degrees.
This gives for $c$ colors the two closed forms ($n$ even and $n$ odd)
$$\frac{1}{4}
(c^{5n^2} + c^{5n^2/2} + 2 c^{5n^2/4}).$$
and
$$\frac{1}{4}
(c^{5n^2} + c^{(5n^2-1)/2+1} + 2 c^{(5n^2-1)/4+1}).$$
A: You have to use Burnside's Lemma. For each of the four possible rotations which are symmetries of the cross, namely rotation by $0^\circ,90^\circ, 180^\circ$ and $270^\circ$, count the number of colorings which are unchanged (fixed) by that rotation. The average of these four numbers is your answer. 


*

*For $0^\circ$ rotation, all of the $c^{5n^2}$ colorings are fixed by this rotation (it leaves everything unchanged).

*For $180^\circ$ rotation, divider the little squares into pairs which are opposite each other through the central square. In order for a coloring to be fixed by this rotation, any two squares in a par must be colored the same. Therefore, there are only $c^{\lfloor 5n^2/2\rfloor +1}$ colorings fixed by $180^\circ$ rotation.

*For $90^\circ$ and $270^\circ$ rotation, squares break into groups of four which all must have the same color, so these each have $c^{\lfloor 5n^2/4\rfloor+1}$ fixed colorings. 
Therefore, the answer is
$$
\frac14\left(c^{5n^2}+c^{\lfloor 5n^2/2\rfloor +1}+2c^{\lfloor 5n^2/4\rfloor+1}\right).
$$
