We will prove the following lemma

Lemma. Consider positive integer $m = 2k+1$ with $k \ge 2$, and the square table of size $m \times m$ in which each cell is filled by one of $n$ colors. Then the number of different ways to color (not duplicated by the rotation) is equal to
$$\frac{n(a^4 + a^2 + 2a)}{4} \quad \text{with } a = n^{k^2+k}.$$
Proof. Note that the middle cell, with symbol C as shown in the image, is not affected by the rotation so there is always $n$ ways to fill it.

Consider the collection of 4 squares $k \times k$ in the corners as set $A = A_1 \cup A_2 \cup A_3 \cup A_4$ and 4 rectangles $1 \times k$ as set $B = B_1 \cup B_2 \cup B_3 \cup B_4$ in such a way that $B_t$ is immediately followed by $A_t$ for $1 \le t \le 4$. We consider 4 pairs consists of two subsets $(A_t, B_t)$ of $A, B$. The number of cells on each pair $(A_t, B_t)$ are $k^2+k$, so the number of ways to fill in each pair of subsets is $a = n^{k^2+k}$. Here, we consider each pair like that as a vertex of some square $XYZT$. We will count the number of coloring for the vertices $X, Y, Z, T$ so that they do not duplicate by the rotation. We have the following cases

(1) All vertices are filled with the same color, there are $a$ ways.

(2) The coloring is alternative, so we just concern how to fill in 2 adjacent vertices with the number of ways is $\frac{a^2-a}{2}$.

(3) Otherwise, there is some identical pairs that is not cyclical of 2, then the number is $a^4-(a^2-a)-a$. This way of coloring has a circular permutations and will generates $\frac{a^4-a^2}{4}$ different ways.

In total, the number of coloring is
$$\frac{a^4 - a^2}{4} + \frac{a^2 - a}{2} + a = \frac{a^4 + a^2 + 2a}{4}.$$

Back to the problem, Using the above lemma when $2k+1=5$, we have a number of ways to fill in the paper $5 \times 5$ that not duplicate each other by the rotation is
$$n \left( \frac{n^{24} + n^{12} + 2n^{6}}{4} \right) = \frac{n^{25} + n^{13} + 2n^{7}}{4}.$$
Denote $S$ as a set of these coloring ways. In this problem, we also need to consider the reflection transformation. We separate $S$ to two types of papers, namely $A$, $B$: papers can and cannot create itself through the vertical and horizontal reflections.

Notice that each paper in $A$ generates 2 different papers in $S$ (should only be counted as 1 in the original problem); meanwhile, each paper in $B$ generates exactly 1 paper in $S$. Hence,
$$2A + B = \frac{n^{25} + n^{13} + 2n^{7}}{4}.$$
Also, it is easy to count $B = n^{15}$ (fill the left half of the piece of paper, also the middle line as well). From there we can calculate
$$2A + 2B = (2A + B) + B = \frac{1}{4}(n^{25} + n^{13} + n^{7}) + n^{15}$$
so
$$A + B = \frac{n^{25} + 4n^{15} + n^{13} + n^{7}}{8}.$$
This is the number of different pieces of paper we need to count. The problem is completely solved. □