imc

Notice that we can rewrite $N^{16}$ as $(N^4{)}^4$. By Fermat's Little Theorem, we know that $N^{(5-1)}\equiv 1(\mathrm{mod}5)$ if $N\not\equiv 0(\mathrm{mod}5)$. Therefore for all $N\not\equiv 0(\mathrm{mod}5)$ we have $N^{16}\equiv (N^4{)}^4\equiv 1^4\equiv 1(\mathrm{mod}5)$. Since $1\le N\le 2020$, and $2020$ is divisible by $5$, $\frac{1}{5}$ of the possible $N$ are divisible by $5$. Therefore, $N^{16}\equiv 1(\mathrm{mod}5)$ with probability $1-\frac{1}{5},$ or $\boxed{\frac{4}{5}}$.