(This started as a comment, but got too long. Maybe someone will find this reasoning helpful.)
The thing is to understand that you can express any rigid system as:
$$\sum {m_i\cdot \vec{r_i}} = \vec{r_c}\sum {m_i} = {M_c}\vec{r_c} = \vec {M_c}$$
Where $({m_i}, \vec{r_i})$ are constituent material points.
(The only special thing about center of mass is that origin point is chosen that $\vec{r_c} = 0$ but it doesn't change any conclusions.)
This has a very nice property of being additive. So you can split your system into any sub-sums of sub-systems, so you can combine material points into larger rigid subsystems. E.g:
$$\vec {M_c} = M_{woman} \cdot \vec r_{woman} + M_{canoe} \cdot \vec r_{canoe}$$
Also, if you apply some force to it, because of additiveness of derivatives and intergrals you can write:
$$\vec{F_{net}} \cdot \vec {M_c} = \sum {m_i\cdot \vec{r_i} \cdot \vec{F_i}}$$
Above means that if you have a combination of some "small forces" acting on your constituants (or sub-systems) you can write it down as some net force acting on center of mass. Now, if you assume there is no net force for the whole system described by $\vec {M_c}$, then by definition it can't move. That is very definition of force. Non-zero force causes displacement. This works both ways. $\Delta \vec {M_c} = 0$ means there was no force, no force means center of mass stayed the same.
Conversely for work you get:
$$W = \vec{F_{net}} \cdot \Delta\vec {M_c} = \sum {m_i\cdot \Delta\vec{r_i} \cdot \vec{F_i}}$$
To move center of mass you would need to spend internal energy or have an external net force to do some work on the system.
But sub-systems can move! $\vec{F_i}$ do not have to be 0.
So after application and work done by force you'll have:
$$\vec {M_c} = \vec {M'_c} = M_{woman} \cdot \vec r'_{woman} + M_{canoe} \cdot \vec r'_{canoe}$$
You could even say, that woman did some work on the canoe... and canoe on the woman with the opposite sign, but energy of their whole system did not change.
The beauty of it is that you absolutely don't care how many exact forces worked on which parts of the system. You can wrap it up as some net force working on center of mass, and because of additiveness you can split it into subs systems. So as long as the woman and canoe look exactly the same in both pictures, it doesn't even matter what shapes they took in between.
You couldn't reverse argument a bit and say that since $$M_{woman} \cdot \vec r_{woman}+ M_{canoe} \cdot \vec r_{canoe} - (M_{woman} \cdot \vec r'_{woman}+ M_{canoe} \cdot \vec r'_{canoe}) = 0$$ i.e. their center of mass did not move (even though their configuration changed) you conclude there was no external force working on them.
