Here is a proof of the generalization of your Weak conjecture to the ring $\mathbf{Z}/m\mathbf{Z}$ where $m$ is any odd positive integer. First let me clarify what is being proved. Let $S$ be the subgroup of $(\mathbf{Z}/m\mathbf{Z})^\times$ which is generated by $2$, and consider a directed graph whose vertices are the sets $gS$ with $g\in\mathbf{Z}/m\mathbf{Z}$ (I emphasize that $g$ need not be in $(\mathbf{Z}/m\mathbf{Z})^\times$), and which has a directed edge from $gS$ to $(3g+1)S$ for every $g\in\mathbf{Z}/m\mathbf{Z}$. I will show:
1) If $3\nmid m$ then this directed graph is strongly connected (in the sense that there is a directed path from any vertex to any other vertex).
2) If $3\mid m$ then for any $g,h\in(\mathbf{Z}/m\mathbf{Z})^\times$ there is a directed path from the vertex $gS$ to the vertex $hS$ (although this path might go through vertices of the form $iS$ with $i\notin (\mathbf{Z}/m\mathbf{Z})^\times$).
Both of these are special cases of the Theorem proved below.
Lemma. Let $n$ be a positive integer, and let $M$ be a subset of $\mathbf{Z}/n\mathbf{Z}$ which contains elements $u,v$ for which $u-v\in(\mathbf{Z}/n\mathbf{Z})^\times$. Then, for every sufficiently large positive integer $s$, every element of $\mathbf{Z}/n\mathbf{Z}$ is the sum of exactly $s$ elements of $M$.
Proof. Let $M_r$ be the set of all sums of exactly $r$ elements of $M$, so that $M_1=M$. Pick $u,v\in M_r$ for which $u-v\in(\mathbf{Z}/n\mathbf{Z})^\times$. Plainly $M_{r+1}$ contains $M_r+u$ and $M_r+v$, so that $\#M_{r+1}\ge\#M_r$. If $\#M_{r+1}=\#M_r$ then we must have $M_r+u=M_r+v$, or equivalently $M_r+(u-v)=M_r$. But then $M_r+\ell(u-v)=M_r$ for any positive integer $\ell$, whence $M_r+w=M_r$ for any $w\in\mathbf{Z}/n\mathbf{Z}$, so we must have $M_r=\mathbf{Z}/n\mathbf{Z}$. Thus, for every $r>0$, either $\#M_{r+1}>\#M_r$ or $M_r=\mathbf{Z}/n\mathbf{Z}$. Since $\#M_r\le m$ for every $r$, it follows that there is some $r$ for which $M_r=\mathbf{Z}/n\mathbf{Z}$, whence $M_s=\mathbf{Z}/n\mathbf{Z}$ for every $s\ge r$. This finishes the proof of the Lemma.
Theorem. Write $m=3^kn$ where $k\ge 0$ and $3\nmid n$. Pick any $g,h\in\mathbf{Z}/m\mathbf{Z}$, and suppose that either $k=0$ or $3\nmid h$.
Then there is a directed path from $g\langle 2\rangle$ to $h\langle 2\rangle$.
Proof. Apply the Lemma with $M$ being the subgroup of $(\mathbf{Z}/n\mathbf{Z})^\times$ generated by $2$ (and with $u=2$ and $v=1$) to obtain a corresponding $s$. Let $r$ be the order of $3$ in $(\mathbf{Z}/n\mathbf{Z})^\times$. Let $\ell$ be the smallest integer such that $\ell>k/r$. Recall that every element of $U:=(\mathbf{Z}/3^k\mathbf{Z})^\times$ is a power of $2$. Since an element $i\in \mathbf{Z}/3^k\mathbf{Z}$ lies in $U$ if and only if $i+3/2$ lies in $U$, it follows that every element of $U$ can be written as $-3/2+2^t$ for some positive integer $t$. The hypothesis that either $k=0$ or $3\nmid h$ implies that the image of $h$ in $\mathbf{Z}/3^k\mathbf{Z}$ is actually in $U$, so there is some $t>0$ for which $h\equiv -3/2+2^t\pmod{3^k}$. Let $b_0,\dots,b_{s-1}$ be nonnegative integers for which $h-3g+s-2^t\equiv\sum_{i=0}^{s-1} 2^{b_i}\pmod{n}$. Define $$a_0:=t,$$
$$a_i:=0 \,\,\text{if either $0<i<r\ell$ or $r\nmid i$},$$
$$a_{r(\ell+j)}:=b_j \,\,\text{for $0\le j\le s-1$}.$$
Then
$$x:=3^{r(\ell+s-1)+1}2^0g+\sum_{i=0}^{r(\ell+s-1)} 3^i 2^{a_i}$$
equals
$$3^{r(\ell+s-1)+1}g + \sum_{i=0}^{r(\ell+s-1)} 3^i + \sum_{j=0}^{s-1} 3^{r(\ell+j)} (2^{b_j}-1) + (2^t-1)$$
$$=3^{r(\ell+s-1)+1}g + \frac{3^{r(\ell+s-1)+1}-1}{3-1} + \sum_{j=0}^{s-1} 3^{r(\ell+j)} (2^{b_j}-1) + (2^t-1).$$
Here $x$ is congruent mod $3^k$ to
$$-\frac12+(2^t-1)=2^t-\frac32=h.$$
Next, $x$ is congruent mod $n$ to
$$3g+\frac{3-1}{3-1}+\sum_{j=0}^{s-1} (2^{b_j}-1) + (2^t-1)=
3g+1+(h-3g+s-2^t)-s+(2^t-1)=h.$$
Therefore $x=h$.
An easy induction shows that the union of the sets $v\langle 2\rangle$ for which the vertex $v\langle 2\rangle$ can be reached from $g\langle 2\rangle$ via a directed path of length $z$ is $\{3^z2^{d_z}g +\sum_{i=0}^{z-1} 3^i 2^{d_i}\colon d_i\in\mathbf{Z}\}$. Thus there is a directed path of length $r(\ell+s-1)+1$ from $g\langle 2\rangle$ to $h\langle 2\rangle$. This finishes the proof of the Theorem.
Best Answer
I'll write $\longrightarrow$ for some standard iterations and $\implies$ for some possibly non-standard iterations. The relaxed Collatz conjecture is that for all $x$, $x \implies 1$. I'll call counterexamples to the conjecture escapees.
First of all, since it's a good example of the notation, a lemma: for all $a$, $4 a \implies 9 a + 1$.
Proof: $4a \implies 12 a + 1 \longrightarrow 36a + 4 \longrightarrow 18a + 2 \longrightarrow 9 a + 1$.
$\square$
Now here is what I will try to show:
Theorem: If $x$ is the smallest escapee, then $4x \implies 1$.
Proof: By elementary considerations, $x \equiv 3 \pmod{4}$. If $x \equiv 2 \pmod{3}$, then $\frac{2 x - 1}{3} \longrightarrow 2 x \longrightarrow x$. So in this case, $2 x$ can't be an escapee, and neither can $4 x$.
The next case is $x \equiv 0 \pmod{3}$. Write $x = 12k+3$. Using the lemma, $4x \implies 27k+7$. Observe that $x > 9k+2 \rightarrow 27k+7$ if $k$ is odd. Now if $x \equiv 15 \pmod{24}$, we also have that $4x \implies 1$.
It's also possible that $x \equiv 3 \pmod{24}$. In that case, we have $x = 24 j + 3$ and $x \longrightarrow 27j+4$. This means $j$ cannot be even. So we drill down again, with $x = 48i + 27 \longrightarrow 81i+49$. This means $i$ cannot be odd. But also, $4x = 4(48i+27) \implies 81i + 92$, so if $i$ is even, the next step brings us under $x$. This covers all the $x \equiv 0 \pmod{3}$ cases.
So far, our result is that if $x \implies 4x$, then $x \equiv 7 \pmod{12}$. Now I'll handle that case. Suppose $x \implies 4x$ and $x = 24k + 7$. Then we have:
$4x = 4(24k+7) \implies 9(24k+7)+1 = 216k+64 \longrightarrow 108k+32 \longrightarrow 54k+16$.
But also, $18k+5 \longrightarrow 54k+16$. This means $18k+5$ is an escapee too, but it can't be, since it's less than $x$. So the hypothesis is false and we have $x = 24k + 19$ instead.
Now, consider:
$4x = 4(24k+19) \implies 216k + 172 \longrightarrow 108k + 86 \longrightarrow 54k + 43 \longrightarrow 162k + 130 \longrightarrow 81k+65$.
By similar reasoning, $k \neq 3 \pmod{4}$, or else we can continue the chain two more steps to get an escapee less than $x$. Finally,
$x = 24k+19 \longrightarrow 72k+58 \longrightarrow 36k + 29 \longrightarrow 108k + 88 \longrightarrow 54k + 44 \longrightarrow 27k + 22$.
shows that $k$ can't be even, and substituting $k = 4 j + 1$:
$27k + 22 = 108j + 49 \longrightarrow 324j + 148 \longrightarrow 162j + 74 \longrightarrow 81j + 37 < x$
completes the proof, since that's all the cases.
$\square$
Corollary: if the Collatz conjecture is false and the smallest counterexample leads back to itself then that number is not a counterexample to the relaxed Collatz conjecture. In other words, if $y$ is the smallest number satisfying $\neg(y \longrightarrow 1)$, then $y \longrightarrow y$ implies $y \implies 1$. This is a weaker version of what I originally thought I had proved.
Corollary: If $4x$ is an escapee, then there is an escapee smaller than $x$. This suggests to me a recursive descent strategy where we attempt to show $z \implies 4k < 4x$. But it's not clear if number-crunching will get us anywhere further.