There would be $2$ such permutations i.e. $(1,2,...,16)$ and $(16,15,...,1)$

Your question is nothing but to find the number of strictly increasing or strictly decreasing functions $f: \{1,2,…,16\} \rightarrow \{1,2,…,16\}.$

$\implies$ Number of strictly increasing functions $f: \{1,2,…,n\} \rightarrow \{1,2,…,m\}$ is $\binom{m}{n}$ and

Number of strictly decreasing functions $f: \{1,2,…,n\} \rightarrow \{1,2,…,m\}$ is $\binom{m}{n}.$

Here, $m=n=16$ and so, Number of strictly increasing functions = $\binom{16}{16} = 1$ and Number of strictly decreasing functions = $\binom{16}{16} = 1$ and so total = $2$

Question would be interesting if they replace “bigger” with “bigger or equal” and lesser with “lesser or equal” (probably, you would not call it "permutation") then you need to find :

$\implies $ Number of nondecreasing functions $f: \{1,2,…,n\} \rightarrow \{1,2,…,m\}$ is $\binom{m+n-1}{m-1}$ and

Number of nonincreasing functions $f: \{1,2,…,n\} \rightarrow \{1,2,…,m\}$ is $\binom{m+n-1}{m-1}.$

So, in this case $(m=n=16)$, Number of nonincreasing or nondecreasing functions $f: \{1,2,…,16\} \rightarrow \{1,2,…,16\}$ is:

$\binom{16+16-1}{16-1} + \binom{16+16-1}{16-1} \ – \ 16$. The subtraction of $16$ denotes there are $16$ permutations which are both nonincreasing and nondecreasing i.e. $(1,1,…,1),(2,2,…,2),…(16,16,…,16).$

Finding formula for such strictly increasing or nondecreasing functions has already been asked in both ISI and GATE exams. So, you could see those derivations for such formulae.