We have the following theorem:
Theorem. Let $p$ be a prime and $G$ be a finite group whose order is divisible by $p$. Then the number of Sylow-$p$ subgroups of $G$ is congruent to $1$ modulo $p$.
We are going to find a counterexample to this theorem.
Let $G=SL_2(F_3)$, where $F_3$ denotes the finite field of order $3$. Then $|G|=24$. Let $n_3$ denote the number of Sylow-$3$ subgroups of $G$. Then the number of elements of order $3$ in $G$ is $2n_3$.
Let $M\in G$ be a matrix of order $3$. Then $M$ satisfies the polynomial $x^3-1=(x-1)^3$. Thus the minimal polynomial of $M$ is either $x-1$, or it is $(x-1)^2=x^2+x+1$. But the minimal polynomial cannot be $x-1$ since the order of $M$ is $3$. So the minimal polynomial of $M$ has to be $x^2+x+1$. Since the characteristic polynomial of $M$ has degree $2$, we see that the minimal and the characteristic polynomial of $M$ coincide, and thus $M$ is a cyclic. This means that there is a vector $v\in F_3^2$ such that $\{v, Mv\}$ is a basis of $F_3^2$. So $M$ is similar to the matrix $N:=\begin{bmatrix} 0 & -1 \\ 1 & -1\end{bmatrix}$.
What we have shown is that all matrices of order $3$ in $G$ are similar to $N$. It is clear that all matrices similar to $N$ are of order $3$.
So the set of all the elements of order $3$ in $G$ is same as the conjugacy class of $N$.
So now we set out to find the size of the conjugacy class of $N$. This is same as the index of the centralizer $C_G(N)$ of $N$. Let $A=\begin{bmatrix} a & b\\ c & d\end{bmatrix}$ be in $C_G(N)$. Then using $ANA^{-1}=N$, we have $b=-c$ and $d=a+b$, and using $\det(A)=1$ we have $a^2+ab+b^2=1$. The solutions to this are $(a, b)=(0, 1), (0, 2), (1, 0), (1, 2), (2, 0), (2, 1)$. So $|C_G(N)|=6$. Therefore $\text{conj}(N)=|G:C_G(N)|=4$.
So we have shown that there are $4$ elements of order $3$ in $G$, which yields there are $2$ Sylow-$3$ subgroups in $G$.
This contradicts the theorem quoted in the beginning of the theorem.
Can somebody please point out my mistake? I know Sylow theorems are not wrong :)
My suspicion (see the comments) turned out to be correct. The centralizer of $N$ in $H=GL_2(\Bbb{F}_3)$ has the same six elements as $C_G(N)$. The OP already showed that $$ X=\left(\begin{array}{cc}a&b\\c&d\end{array}\right) $$ commutes with $N$, iff $c=-b$ and $d=a+b$, and then listed the six such matrices with determinant $a^2+ab+b^2=1$. The field $\Bbb{F}_3=\{0,\pm1\}$, so we want to check whether there are matrices $X$ that commute with $N$ and have determinant $=-1$. But the equation $$ a^2+ab+b^2=-1 $$ has no solutions $(a,b)\in\Bbb{F}_3^2$. This is because modulo $3$ we have $$ a^2+ab+b^2=a^2-2ab+b^2=(a-b)^2, $$ and $-1$ is not a quadratic residue modulo $3$. This means that $\det X\neq-1$ for all $X\in C_G(N)$.
Therefore $C_H(N)=C_G(N)$ has six elements, and the conjugacy class of $N$ in $H$ has eight elements. Because $G\lhd H$, the conjugacy class is contained in $G$, but must split into two conjugacy classes of $G$.
To give a more complete picture I will show that the matrices $$ N_1=\left(\begin{array}{cc}1&1\\0&1\end{array}\right) $$ and $$ N_2=\left(\begin{array}{cc}1&-1\\0&1\end{array}\right) $$ are conjugate in $H$ but not in $G$. With $X$ as above the equation $XN_1=N_2X$ is equivalent to $$ c=0\qquad\text{and}\qquad a+b=b-d. $$ This has a solution $1=a=b=-d$, but also forces $d=-a$ whence $\det X=-a^2=-1$.
In general matrices of the form $\pmatrix{1&a\cr0&1\cr},a\neq0,$ are all conjugate in $GL_2(\Bbb{F}_p)$. But they split into two conjugacy classes in $SL_2(\Bbb{F}_p)$ according to whether $a$ is a quadratic residue or a quadratic non-residue modulo $p$.
It may be worth observing that we see similar occasional splitting of conjugacy classes of the symmetric group $S_n$ (i.e. permutations of identical cycle types) into two classes in the index two subgroup $A_n$. The mechanism for detecting when this splitting happens is the same we just saw here - study the size of the centralizer in both the bigger group and the subgroup of interest.
If the centralizer of an even permutation $\sigma$ only contains other even permutations, then the conjugacy class is split into two equal size conjugacy classes of $A_n$, because the size of the conjugacy class is given by the index of the centralizer. OTOH, if a permutation $\sigma\in A_n$ is centralized by at least one odd permutation (in which case half the permutations in the centralizer will actually be odd), then $C_{A_n}(\sigma)$ will be of index two in $C_{S_n}(\sigma)$. Consequently the size of the conjugacy class will remain the same, and hence the class won't split.