Let $f(n)$ be A000045(n), i.e., Fibonacci numbers: $f(n)=f(n-1)+f(n-2)$ for $n>1$ with $f(0)=0$ and $f(1)=1$.
Let $g(n)$ be A072649, i.e., $n$ occurs $f(n)$ times. The sequence begins with $$1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7$$ Here we also consider that $g(0)=0$.
Let $$h(n)=n-f(g(n)+1)$$ Claude Chaunier pointed out in his comment that
From OEIS we see $f(g(n)+1)$ is the highest Fibonacci number $\leqslant n$ and $h(n)$ is how far $n$ is from it.
So the binary analogs of these functions are: powers of $2$ for $f(n)$, $\left\lfloor\log_2 n\right\rfloor$ for $g(n)$ and $n$ without the most significant bit for $h(n)$.
Let $a(n)$ be an integer sequence such that $a(0)=0$, $a(n)=n$ if $n$ is Fibonacci number, otherwise $$a(n)=n - f(g(h(n)-1)) + [g(n)\operatorname{mod}2 = g(h(n)-1)\operatorname{mod}2]\cdot f(g(n)-2)$$ I conjecture that $a(n)$ is a permutation of the nonnegative integers.
Here is the PARI prog to verify this conjecture:
g(n)=local(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2) \\ from A072649
a(n) = if(n>0, my(A=g(n), B=n-fibonacci(g(n)+1), C=g(B-1)); n + if(B>0, - fibonacci(C) + if(A%2==C%2, fibonacci(A-2))))
test(n) = my(A=0); while(!(a(A)==n), A++); A
Is there a way to prove it? Is there a binary analog of this permutation?