Flow graph that requires pushing back flow in Ford Fulkerson Does there exist a flow graph that always requires flow to be pushed back no matter what ordering of augmenting paths is chosen in Ford Fulkerson?

Let's assume we use the standard procedure of repeating this step:


Find an augmenting path $p$ in residual graph $G_R$ of $G$.
Let $c$ be the minimum capacity edge in $p$.
Increase the flow on every edge in $p$ by $c$.
Update $G_R$.


The key here is step 1, where select augmenting paths. For many graphs, if we had an oracle to tell us which augmenting paths to use, we would never need to push flow back. I'm am curious if there is a case for which, regardless of augmenting paths and their orders, we will always be required to "push flow back". To clarify what I mean:


  To push flow back in $G$, means to increase the flow on an edge $(u,v)$ in $G_R$ such that edge $(v,u)$ exists in $G$.


If this is not possible, I would also be interested in a proof that such ordering of augmenting paths always exists? If it is possible, does it generalize to any number of nodes $n$? This question is alluded at in the ending sentences of this answer, but provides no proof. 

My initial thoughts were that this would be a trivial proof. However, there are many times when the optimal flow along an augmenting path may not be equivalent to its minimum capacity edge. Since (by step 3) we require paths to be filled to their minimum capacity, we cannot easily meet this. My next thought would be that there should exist at least one augmenting path such that its max flow is equivalent to its minimum capacity edge. This is obvious by the Max flow min cut Theorem, but I am not sure how this would apply to the proof. With this, we may be able to get an inductive proof that it is always possible, but I am really unsure of this strategy as well. Any help would be appreciated.
I think it is possible to reason backwards from an optimal solution$^1$ to the order in which the paths have to be augmented to not have any pushing-back:

In the solution, for any path from the source to the sink, where the flow is larger than zero, there is at least one edge, that constrains that flow, i.e: its flow is at capacity.
If it were not, there would have been an augmenting path along that path, i.e. the solution would not be optimal.
Also, of the edges that are at capacity, there has to be one (or multiple) with the lowest possible capacity. Furthermore, this edge has to be part of some path between source and sink where each edge has a flow at least as high. 
We take such a path, call it $P_1$ and remove its 'flow capacity' from the initial graph (i.e. reduce the capacity of the edges along the path).
We repeat this with the remaining graph (resulting in $P_2, P_3, \ldots, P_n$) until there is no path between source and sink remaining, where all edges have a non-zero capacity.

Then, $P_1, P_2, \ldots, P_n$ is the order in which the paths have to be chosen s.t. there is never any pushing back necessary. 

$1$: This only holds for optimal solutions that have been found with the FF algorithm. E.g. in a graph where 5 edges with capacity 2 go into node X and only one edge leaves it with capacity 5, there would be a max flow where each of the 5 edges has a flow of 1/2 and the outgoing has a flow of 5/5. Then the 5/5 edge would be the lowest edge at capacity but there is no path from source to sink in that graph where each edge has a flow of 5 or more. This is no problem for the algorithm outlined above because such a max-flow solution would not be found with FF.