Thoughts on the Collatz conjecture; integers added to powers of 2

The following part of the argument seems to be problematical.

"Now suppose we have some number n which definitely goes to $1$. We now perform the Collatz iteration on $2^{a}+n$. The parity of this number is identical to the parity of $n$.
If we were to perform $3n+1$, then we would have $3*2^{a}+3n+1$
Indeed, since the parity of the number is determined by $n$, the iteration is identical to that of $n$."

The first example I tested had $n=3$ and $2^a=8$. So $2^a+n=11$.

The Collatz sequence with first entry $3$ goes $$3, 10, 5, 16, 8, 4, 2, 1.$$

The Collatz sequence with first entry $11$ goes $$11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.$$

The iterations are not identical.

It is true that if $a$ is fairly large, and $n$ is not, then the patterns of odd and even will, for a while, be the same. In our example above, they were the same until the sixth number.

If I got your thinking correctly, then the key, or better:the general aspect, for your observation is the following transfertable (where $k$ is a nonnegative integer): $$ \small \begin{array}{r|rrcrr||r|rrcrr} \text{class} &a&&\to&b&&\text{class}&a&&\to&b\\ \hline \\ 1& 2^1 \cdot 2k &+ 3 &\to& 3 \cdot 2 k &+5 & 2& 2^2 \cdot 2k &+ 1 &\to& 3 \cdot 2 k &+1 \\ 3& 2^3 \cdot 2k &+ 13 &\to& 3 \cdot 2 k &+5 & 4&2^4 \cdot 2k &+ 5 &\to& 3 \cdot 2 k &+1 \\ 5& 2^5 \cdot 2k &+ 53 &\to& 3 \cdot 2 k &+5 & 6&2^6 \cdot 2k &+ 21 &\to& 3 \cdot 2 k &+1 \\ ...&... &&\to&...&& & ...&&\to&...\\ A& 2^A\cdot 2k&+{5 \cdot 2^A-1\over3} & \to & 6k&+5& B& 2^B\cdot 2k&+{ 2^B-1\over3} & \to & 6k&+1& \end{array}$$ where $A\in \{1,3,5,7,...\}$ and $B \in \{2,4,6,8,...\}$
So what you seem to have observed is that if you have some (say "residue") $r \in \{3,13,53,213,853,...\}$ or $r \in \{1,5,21,85,341,...\}$ then you can add an associated power of $2$ and the transformation might be called to be of the same "class".
The table above shows that this observation might even be generalized: not only is it possible to add a suitable power of $2$ but even arbitrary many version with $k$'th multiples of that power can be said to "belong to the same class".
Your additional observations involving the domain of the negative numbers are still matched if you introduce negative $k$ in the above table. And moreover: once we allow negative numbers, then it might be interesting, that the residue-class $3$ means also $-1 \pmod 4$ and thus $+1$ and $-1$ form the "trivial cycles" in class $1$ with $k=-1$ and class $2$ with $k=0$