Let the Collatz function go $(3x+1)/2$ for odd numbers and $x/2$ for even.
Now running the function, write a $1$ every time you hit an odd number and a $0$ every time you hit an even number. By this means you can record a function $T$ from your domain into binary numbers. $T(x)$ is sometimes called the Parity Vector of a number's Collatz sequence. For example the number $1$ follows the orbit $1,2,1,2,\ldots$ so it writes the binary number $T(x)=\ldots0101_2=-\frac13$
$T$ is a 2-adic isometry $\Bbb Z_2^\times\to\Bbb Z_2^\times$ having the fixed points $0$ and $-1$.
But if we write a $-1$ instead of a $1$, I think by topological conjugacy we still get an isometry and it still has two fixed points. Zero is obviously still a fixed point, but what is the other fixed point of the new function?
My Attempt
Not part of the question - only provided as evidence of my own attempt in line with site policy. I have a hunch this needs to be attacked by somehow writing an equation with two copies of Newton's method, one on each side, and at each step, asing whether $0$ or $-1$ is the solution to the next digit but I'm unfamiliar with even basic applications of Newton's method.
Edit
I did a bit of algebra and I think seek the number such that $T(x)=-x$, i.e. its Collatz Parity Vector is itself negated. Also, heuristically the numbers $n$ among $-\frac13\Bbb N$ are good candidates, in particular ones satisfying $3n\equiv\{1,2\}\pmod 3$
Let $C(x) = \begin{cases}\frac x2&x\text{ is even}\\\frac{3x+1}2&x\text{ is odd}\end{cases}$ be the Collatz function. The Collatz sequence of $x$ is then the sequence $x, C(x), C^2(x), C^3(x), \dots$, which is the Collatz sequence of $C(x)$, with $x$ prepended. In particular, the Collatz sequence of $2x$ is the Collatz sequence of $x$ with $2x$ prepended. From this, it follows that $T(2x) = 2T(x)$. Multiplying by $2$ just shifts both the bit sequences of $x$ and $T(x)$ to the left by one bit.
Thus if $-T(x) = x$, then $-T(2x) = -2T(x) = 2x$. That is, if $x$ is a fixed point of $-T$, then so is $2x$. So if $-T$ has any fixed points other than $0$, it has infinitely many of them.
It also goes the other way: if $x$ is an even fixed point of $-T$, then $\frac x2$ is a fixed point as well. Thus if $-T$ has non-zero fixed points, one of them will be odd. So suppose that $x$ is an odd solution to $T(x) = -x$. To negate in the 2-adic integers, you find the rightmost $1$ bit, then flip all bits to its left (the "twos-complement"). Suppose $x$ has only finitely many $0$ bits. I.e., it is an ordinary negative integer. Then $-x = T(x)$ will have only finitely many $1$ bits. This means its Collatz sequence is eventually always even. But if $x \ne 0$, this cannot be, as the first entry after all odd entries are finished, would have to be divisible by $2$ infinitely many times, meaning its 2-adic representation is all $0$ bits, and thus that the number is $0$. But $C(y) = 0$ has only one solution: $y = 0$, so this is impossible.
Next suppose that $x$ has only finitely many $1$ bits. I.e., it is an ordinary positive integer. This means that its Collatz sequence is eventually always odd. Again look at the first entry after the last even entry in the sequence. This Collatz sequence for this number will be strictly increasing. Thus its existence would disprove the conjecture. As the conjecture has not been disproven, we can be sure no such number is known. On the other hand, I am no expert on the Collatz conjecture, but as far as I am aware, it has not been proven no such increasing sequence exists. So it is still a possibility.
So if the conjecture is true, then the only possible non-zero fixed points of $-T$ would have to be strictly 2-adic, not in the real integers.