Suppose $A^5=I$ and write $\zeta:=\exp({{2\pi i}\over 5})$. Each eigenvalue is a power of $\zeta$ hence $F_A=(x-\zeta^{k_1})\cdots (x-\zeta^{k_5})$ where all $k_i\in \{0,1,2,3,4\}$. Also $\sum_1^5 \zeta ^{k_i}=0$. The minimum polynomial of $\zeta$ equals $f:=x^4+x^3+x^2+x+1$ hence $f| \sum_1^5 x^{k_i}$. Since RHS has degree at most 4 we have equality and $\{k_1,\cdots,k_5\}=\{0,1,2,3,4\}$. In particular $F_A(1)=0$, contradiction.