This is exercise 12.2.16 of Abstract Algebra by Dummit and Foote.
Show that $x^5-1 =(x-1)(x^2-4x+1)(x^2+5x+1)$ in $\mathbb F_{19}[x]$. Use this to determine, up to similarity, all $2 \times 2$ matrices with entries from $\mathbb F_{19}$ of (multiplicative order) $5$.
First of all, the multiplicative group of $\mathbb F_{19}$ is $C_{18}$, the cyclic group of order 18, with generator $2$, so I presume they mean that the matrices are of order $5$.
Secondly, I prove the decomposition result as stated and then see that due to Cayley-Hamilton we know that any $2 \times 2$ matrix $A$ with minimal polynomial that divides $x^5-1$ must also satisfy $A^5-I=0 \implies A^5= I$.
The candidate polynomials are: $x-1$, $x^2-4x+1$ and $x^2+5x+1$.
This question is similar to what they do on page 487, but I am not quite sure how one goes from these candidates to the list of permissible invariant factors, and after that, the matrices.
We looks at the divisors of $x^5-1$ and use the decomposition as given. Notice for a $2 \times 2$ matrix it is only possible to get $2$ blocks of size $1$ or $1$ block of size $2\times 2$.
Recall that for a polynomial of the form $b_0 +b_1x +x^2$, we get the following companion matrix according to page 475: $$ \begin{pmatrix} 0 & -b_0 \\ 1 & -b_1 \end{pmatrix}$$ A $1\times 1$ block simply has the value of $-b_0$ for $b_0 + x$.
$x-1$, $x-1$ give us: $$\begin{pmatrix} -(-1) & 0 \\ 0 & -(-1) \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}=I.$$ This matrix is of course not of order $5$, but of order $1$, therefore it satisfies $I^5=I$.
$x^2-4x+1$ gives us: $$A= \begin{pmatrix} 0 & -1 \\ 1 & 4 \end{pmatrix}\equiv \begin{pmatrix} 0 & 18 \\ 1 & 4 \end{pmatrix} \bmod {19}$$
$x^2+5x+1$ gives us: $$B=\begin{pmatrix} 0 & -1 \\ 1 & -5 \end{pmatrix}\equiv \begin{pmatrix} 0 & 18 \\ 1 & 14 \end{pmatrix} \bmod {19}$$ One can easily verify that $$ A^5 \equiv B^5 \equiv 1$$ Since $5$ ia prime and the matrices are nonidentity, this is the order.