$A$ is symmetric (or selfadjoint, if your matrices are complex), so it is diagonalizable.
It is not idempotent, because $A^2=5A$.
It is not nilpotent, because $A^n=5^{n-1}A$.
The minimal polynomial of $A$ is $p_m(t)=t(t-5)$, while the characteristic polynomial is $p_c(t)=t^{4}(t-5)$. So they are different.
The characteristic polynomial can be obtained this way: we have $ A=\begin{bmatrix}1&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 1&0&\cdots&0 \end{bmatrix}\,\begin{bmatrix}1&\cdots&1\\0&\cdots&0\\ &\ddots\\0&\cdots&0 \end{bmatrix}, $ so its spectrum agrees with that of $ \begin{bmatrix}1&\cdots&1\\0&\cdots&0\\ &\ddots\\0&\cdots&0 \end{bmatrix}\,\begin{bmatrix}1&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 1&0&\cdots&0 \end{bmatrix}=\begin{bmatrix}5&0&\cdots&0\\0&0&\cdots&0\\ & & \ddots \\0& & \cdots &0 \end{bmatrix}. $ The eigenvalues are then $5,0,0,0,0$, and so $p_c(t)=t^4(t-5)$. As $A^5-5A=0$, the minimal polynomial is at most $t(t-5)$. But it also requires $0$ and $5$ as roots, so $p_m(t)=t(t-5)$.