Count it

First Pile: let $B$ be the number of black cards and $R$ be the number of red cards. So, we have, $\boxed{R=5B}$ $\color{#D61F06}(1)$

Second Pile: let $b$ be the number of black cards and $r$ be the number of red cards. So, we have, $b=16+r$ or $\boxed{r=b-16}$ $\color{#D61F06}(2)$

In a standard deck of $52$ cards, there are $26$ red cards and $26$ black cards. So, we have

$\boxed{R+r=26}$ $\color{#D61F06}(3)$

and

$B+b=26$ $\large \implies$ $\boxed{B=26-b}$ $\color{#D61F06}(4)$

Substitute $\color{#D61F06}(1)$ and $\color{#D61F06}(2)$ in $\color{#D61F06}(3)$ . We have,

$5B+b-16=26$ $\color{#D61F06}(5)$

Substitute $\color{#D61F06}(4)$ in $\color{#D61F06}(5)$ . We have,

$5(26-b)+b-16=26$ $\large \implies$ $130-5b+b-16=26$ $\large \implies$ $88=4b$ $\large \implies$ $\color{#3D99F6}\boxed{\large b=22}$

It follows that,

$r=22-16$ $\large \implies$ $\color{#3D99F6}\boxed{\large r=6}$

Finally,

$b+r=22+6=$ $\color{#20A900}\large \boxed{28}$

Let the numbers of red and black cards in the first pile be $r$ and $b$ respectively. Then the numbers of red and black cards in the second pile are $26-r$ and $26-b$ respectively.

From the first pile: $\color{#3D99F6}r = 5b$ . From the second pile:

$\begin{aligned} 26 - b & = 26 - r + 16 \\ \implies \color{#3D99F6} r & = b + 16 & \small \color{#3D99F6} \text{Note that }r=5b \\ \color{#3D99F6} 5b & = b + 16 \\ \implies b & = 4 \\ \implies r & = 5b = 20 \end{aligned}$

Therefore, the number of cards in the first pile $N_1 = r+b = 20+4 = 24$ , then the number of cards in the second pile $N_2 = 52 - N_1 = 52-24 = \boxed{28}$ .