I thought about this on the walk home from the train today.  I think
we both agree that a rapidly-varying external force can affect whether
the pulling coordinate is a good reaction coordinate.  But what about
a constant force (or even static arbitrary potential) applied along
only along the observed extension coordinate?  Could a static force
applied along a chosen coordinate, if large enough, affect whether the
coordinate is a good reaction coordinate or not?

I wonder if there is a way to demonstrate definitively yes or no.
Turning to the Mori-Zwanzig projection operator formalism, suppose we
have coordinates q and momenta p and wish to project onto some
coordinate A(q).  MZ tells us we can write the correlation function
along A as

C(t) = - \int_0^t dt' M(t') C(t-t')

where

C(t) = (A(t), A*(0))  [correlation function]
M(t) = (F(t), F*(0)) / (A,A)  [memory function]

We know that the external force will probably affect the dynamics, and
hence the correlation function C(t).  But can it really change an
orders-of-magnitude separation between the decay time constant for the
memory function M(t) and C(t)?

Let P be the projection operator onto A, such that PB = (B,A)/(A,A) A;
let Q = 1-P.

The random force F(t) is defined as

F(t) = exp[iQLA]

where L is the Liouvillean for whatever model of dynamics is governing
the microscopic process.

If the process is governed by Hamiltonian dynamics, we can write

L = - Dq H . Dp + Dp H . Dq

where Dq is the gradient operator for q, and Dp the gradient for p.

Adding a linear force in A(q), we have now

H' = H + s A(q)

such that

L' = L - s Dq A . Dp

Now, inserting this into our equation for the random force, we get the
new random force

F'(t) = exp[iQL'A] = exp[iQ(L - s Dq A . Dp)A] = exp[iQLA] = F(t)

so the random force is unchanged.  This is what we would expect, since
the force is applied only along the observed coordinate A(q).

What *is* changed, however, is the averaging in computing the memory function:

M(t) = (F(t),F(0)) / (A,A)

Since the scalar product (A,B) denotes

(A,B) = \int dq \int dp rho(q,p) A(q,p) B*(q,p)

where rho(q,p) is the stationary distribution, we see that the
*integrand* F(t) F*(0) of (F(t),F(0)) isn't changed, but the weighting
measure rho(q,p) will be changed to reflect the shift in population
due to the external biasing force.

So, the big question remains unresolved.  We know that the integrand
doesn't change, but it's possible that the force will shift the
equilibrium such that a different set of "random force trajectories"
are emphasized, causing a change in the apparent timescale of decay of
M(t) relative to C(t).

Anyway, I just thought the analysis was sort of interesting.  At
first, I was excited that the force couldn't affect F(t), but then I
realized that wasn't the whole story...