Functional equation with delta

Is it possible to exist continuous R->R function f such that:
f = delta(f o f) (*)
and f(x) = 0 for every real x is false.
If so is f unique?

Notation:
(delta(f))(x) = f(x+1) - f(x)
(f o g)(x) = f(g(x))

Therefore (*) is equivalent to: f(x) = f(f(x+1)) - f(f(x)).

Thank you,
Darrell H

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