We put a square $3 \text{ x } 3$ from $9$ square tiles in two types:
or
which can be freely rotated (we allow also symmetries). Calculate in how many different ways you can do that if we identify such systems that one goes to the other with some rotation or symmetry of big square.
My solution:
I am going to use polya counting theory
https://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem
$$\begin{array}{|c|c|c|c||c|c|}
\hline
&G \in G & \text{How many}& \text{Id x }\\ \hline
a) &\text{id} & 1 & x_{1}^{9} \\ \hline
b) &\text{90 degree rotation} & 2 & x_1 \cdot x_4^{2} \\ \hline
c) &\text{180 degree rotation } & 1 & x_1 \cdot x_2^{4}\\ \hline
d) &\text{ diagonal symmetry} & 2 & x_{1}^{3} \cdot x_2^{3}\\ \hline
e) &\text{centres symmetry } & 2 & x_1^3 \cdot x_2^{3}\\ \hline
\end{array}$$
We must include our tiles:
a) Here freely, so we have $6$ types: $6^9$
b) 
For each $4$-element cycle we choose whether the first will be:
The rest for him is clearly designated:
The middle one can't be adjusted to fit all $4$ rotations, so: $0$
c) Middle: 
and the rest of the cycles as in b): $2\cdot 6^4$
d) In the middle in $6$ ways, the rest as in b): $6^3 \cdot 6^3=6^6$
e) $6^6$ as in d)
Solution: $$\frac{1}{8} \cdot (2\cdot 6^6 + 2\cdot 6^6 +2\cdot 6^4+0+6^9)$$
Is my solution correct?
Best Answer
Let the small square with 1 shaded triangle be called A and the one with 2 shaded triangles be called B. We shall apply Burnside's Lemma to the group of symmetries of the large square acting on the $6^9$ patterns.
The identity fixes all $6^9$ patterns.
The rotations of 90 and 270 cannot fix the centre square and so fix no patterns.
The rotation of 180 fixes the centre square if and only if it is the square B. We can then have 2 choices for this square. The other 8 squares are in pairs where we have a free choice for one of each pair and therefore the total number of patterns is $2\cdot6^4$.
The two reflections in a diagonal cannot fix a corner square and so fix no patterns.
Finally consider a reflection in a horizontal line. (The result for the reflection in a vertical line will be similar.) The central row of squares must each consist of a square B or a square A in just two of its orientations. The remaining six squares are in pairs where we have a free choice for one of each pair and therefore the total number of patterns is $4^3.6^3$.
The solution is $\frac{1}{8} \cdot (6^9+2\cdot 6^4 + 2\cdot 4^3\cdot6^3)$.