Answer: 177

Small Hint:

Two reflection symmetries compose to a translation: $f(398 - x) = f(2158 - x)$ forces $f(t) = f(t + 1760),$ and similarly $f(t) = f(t + 1056)$

Big Hint:

So $f$ has period $\gcd(1760, 1056) = 352,$ plus the symmetry $f(x) = f(46 - x);$ count residues mod $352$ up to that pairing

Solution:

Since $f(398 - x) = f(2158 - x)$ for all $x,$ substituting $t = 398 - x$ gives $f(t) = f(t + 1760);$ likewise $f(2158 - x) = f(3214 - x)$ gives period $1056.$ Combining, $f$ has period $\gcd(1760, 1056) = 352.$ Reducing $398$ mod $352,$ the symmetry $f(x) = f(398 - x)$ becomes $f(x) = f(46 - x).$

So $f$ is determined by residues mod $352,$ with residues $r$ and $46 - r$ forced to share a value. This pairing has exactly two fixed points, from $2r \equiv 46 \pmod{352}:$ $r = 23$ and $r = 199.$ Hence there are at most $\frac{352 - 2}{2} + 2 = 177$ classes, and since $0, 1, \ldots, 999$ covers every residue mod $352,$ the list contains at most $177$ different values.

This is achievable: $f(x) = \cos\frac{2\pi(x - 23)}{352}$ satisfies all three given symmetries (each of $398,$ $2158,$ $3214$ is $\equiv 46$ mod $352$), and two integers get equal values only when their residues are paired. So the answer is $177.$