Here is the matrix I am trying to find its JCF:
$$ \begin{pmatrix} 1 & 2 & 0 & 0\\ 0 & 1 & 2 & 0\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 1 \end{pmatrix} $$ I found that its characteristic polynomial is $(\lambda - 1)^4$ and hence it has eigenvalue 1 with multiplicity four then the four conjugacy classes corresponding to its minimal polynomial are $$ \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} $$ (corresponding to if the minimal polynomial is $\lambda -1 $) Or $$ \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} $$ (corresponding to if the minimal polynomial is ($\lambda -1 )^2$) Or
$$ \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} $$ (corresponding to if the minimal polynomial is ($\lambda -1 )^3$)
$$ \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1 \end{pmatrix} $$ (corresponding to if the minimal polynomial is ($\lambda -1 )^4$)
But then why should I conclude that the JCF of A is $$ \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1 \end{pmatrix} $$ ? Could anyone clarify this to me or at least tell me what should I do next to arrive to the correct JCF?
From here you should work with geometric multiplicity and minimal polynomial.
Geometric multiplicity
We have to calculate the dimension of the eigenspace relative to $\lambda = 1$, so we have to compute $\dim(Ker(A -\lambda I))$ where $A$ is your matrix. We have $$A -\lambda I = \begin{pmatrix} 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}$$ We can rapidly conclude that this eigenspace is generated by $v = (0, 0, 0, 1)^T$ so the geometric multiplicity (corresponding to the number of Jordan block for $\lambda = 1$) is one $\implies$ one Jordan block. So in this case we end because the canonical jordan form is $$J = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$
Minimal Polynomial
In this case we don't need to investigate any further but if for example geometric multiplicity of $\lambda = 1$ is equal to $2$, the we have to evaluate the minimal polynomial and the algebraic multiplicity is the dimension of the biggest Jordan block. If we want to evaluate minimal polynomial we should consider that shares the same roots of characteristic polynomial. As you point out we have four possibilities
And be the minimal polynomial $p_A(\lambda)$ for our matrix $A$, means that $p_A(A)=0$ (forgive me for this little notation abuse). After some calculation you should obtain that in this case minimal and characteristic polynomial coincide. We conclude that the dimension of the biggest Jordan Block is $4$, as deduced above. Hope this helps!