What the commenter in your linked problem likely meant, is that it doesn't matter where you put the current source, the resistance across that current source will be the same everywhere due to symmetry reasons because all resistors have the same value.
So I think he actually isn't really trying to say what you're trying to do.
In general, what you are trying to do can be considered impossible except in very specific cases. These cases can be deduced as follows:
Consider you can write a reduced form of the nodal analysis: i.e. combining resistors together until you end up with just the following two nodes, while using the common \$V_3\$ node as the ground node:
$$Y\cdot \left(\begin{matrix}
V_{5a} \\
V_{9a}
\end{matrix}\right) = \left(\begin{matrix}
1 \\
0
\end{matrix}\right)$$
Remember that \$R_{a}\$ is actually just \$V_{9a}/1A\$. We don't have access to \$V_{5a}\$ in what you want to do. We are actually interested in the solution of
$$Y\cdot \left(\begin{matrix}
V_{5b} \\
V_{9b}
\end{matrix}\right) = \left(\begin{matrix}
0 \\
-1
\end{matrix}\right)$$
More specifically, we are actually only interested in \$V_{5b}\$, as \$V_{5b}/1A\$ will be the resistance.
Now the easiest would be to just solve the second set of equations (which is what you get if you just analyze it with the current source between 3 and 5). However, what you appear to want to do is start with the solution of the first set of equations, \$(V_{5a}, V_{9a})\$, and then immediately jump to the solution of the second set of equations.
It is possible by realizing that in both situations, the Y-matrix is identical, while the right-hand side is just a permutation, or
$$\left(\begin{matrix}
0 \\
-1
\end{matrix}\right) = \left(\begin{matrix}
0 & 1 \\
-1 & 0
\end{matrix}\right)\cdot\left(\begin{matrix}
1 \\
0
\end{matrix}\right)$$
This allows us to link both of these sets of equations together, such that you find that:
$$\left(\begin{matrix}
V_{5b} \\
V_{9b}
\end{matrix}\right) = \left(Y^{-1}\cdot\left(\begin{matrix}
0 & 1 \\
-1 & 0
\end{matrix}\right)\cdot Y\right)\cdot\left(\begin{matrix}
V_{5a} \\
V_{9a}
\end{matrix}\right) = F'\cdot\left(\begin{matrix}
V_{5a} \\
V_{9a}
\end{matrix}\right)$$
(\$F\$ contains unit-less values)
Now, saying that we could determine \$V_{5b}\$ only from knowing \$V_{9a}\$ implies that \$F'_{11} = 0\$, because else we would need to know \$V_{5a}\$ as well, since \$V_{5b} = F'_{11} V_{5a} + F'_{12} V_{9a}\$. It also means that you have a way to calculate \$F'_{12}\$, which is maybe not as straightforward as you might like. We then have that:
$$F'_{11} = 0 \Rightarrow Y_{22}\cdot Y_{21} + Y_{11}\cdot Y_{12} = 0$$
$$F'_{12} = \frac{Y_{12}^2 + Y_{22}^2}{Y_{11}Y_{22}-Y_{12}Y_{21}}$$
I think at this point that you can see that this case is not obviously satisfied. In your specific case, it will even never happen without a lot of open-circuits. In your example, the Y-matrix can be reduced to
$$Y = \left(\begin{matrix}
g_{3-5}+g_{5-9} & -g_{3-9} \\
-g_{3-9} & g_{3-9}+g_{5-9}
\end{matrix}\right)$$
If all \$g\$ are positive, then \$F'_{11} \neq 0\$. The only exception would be a situation where \$g_{5-9}=0\$, or in other words, that there is no conductive path between the both of them. In that case, \$F'_{12} = \frac{g_{3-9}}{g_{3-5}}\$ and this is trivial because it basically requires you to know the resistance \$g_{3-9}\$...
