The question "can a given homotopy idempotent admit multiple inequivalent coherentifications" ought to be approachable by the standard spectral sequence machinery, so let me try to do that. I'll more or less follow the Dwyer-Kan approach.

Let $M=\langle f\,|\, f^2=f \rangle = \{1,f\}$, the walking idempotent as a monoid. Let $\mathcal{S}=$ the simplicially enriched category of CW-complexes, and $h\mathcal{S}$ its homotopy category.

Clearly, a "homotopy idempotent" gives rise to a functor $M\to h\mathcal{S}$. (But not quite conversely, as you've built a choice of homotopy into the definition.) A "coherentification" of it should amount to a commutative diagram
$$
\begin{array}{ccc}
{\widetilde{M}} & \to & \mathcal{S} \\ \downarrow &&\downarrow \\ M & \to & h\mathcal{S},
\end{array}
$$
where $\widetilde{M}\to M$ is a cofibrant approximation to $M$ in simplicial monoids.

Let's fix a space $X$ and $f\colon X\to X$ such that $ff=f$, thus determining a functor $\gamma\colon \widetilde{M}\to M\to \mathcal{S}$. We want to compute something about a full subspace of
$$\newcommand{\Map}{\mathrm{Map}}
\Map_{s\text{Monoids}}( \widetilde{M}, \Map(X,X) )
$$
whose points are $\phi\colon \widetilde{M}\to \Map(X,X)$ such that $h(\phi)\colon \pi_0\widetilde{M}\to \pi_0\Map(X,X)$ coincides with our chosen $\gamma$.

There's a spectral sequence for this:
$$
E_2^{s,t}= H_Q^s(M, A_t) \Longrightarrow \pi_{t-s}\Map_{s\text{Monoids}}(\widetilde{M},\Map(X,X))_{\gamma}.
$$
The corner $E_2^{0,0}$ is anomalous. It is not given by cohomology; rather, it corresponds to a choice of $\gamma\colon M\to \pi_0\Map(X,X)$, which we have fixed here. We are going to be interested in the groups $E^{t,t}_2=H_Q^t(M,A_t)$ for $t\geq 1$, which potentially contribute to $\pi_0$.

The cohomology is "Quillen cohomology" of the monoid $M$. The coefficients are in an abelian group object in $\text{Monoids}_{/M}$. Such an abelian group object amounts to:

- For each $x\in M$, an abelian group $A(x)$.
- For each $x,y\in M$, group homomorphisms $x\cdot A(y)\to A(xy)$ and $\cdot y\colon A(x)\to A(xy)$ which are natural (e.g., $x\cdot(y\cdot a)=(xy)\cdot a$ and $1\cdot a=a$, and similarly on the right), and which commute: $(x\cdot a)\cdot y= x\cdot (a\cdot y)$.

From this you can build the monoid $A := \coprod A(x)$ with product defined by
$$
a\cdot b := (a\cdot y)+ (x\cdot b) \qquad\text{for $a\in A(x), b\in A(y)$.}
$$

For a monoid $M$ acting on a space $X$, we use the coefficient systems defined by
$$
A_t(f) := \pi_t \Map(X,X)_f \qquad \text{for $f\in M$.}
$$
The "$M$-actions" are defined by pointwise pre-or-post composition of a map $a\colon S^t\to \Map(X,X)$ with the constant map $S^t\to *\xrightarrow{f} M\to \Map(X,X)$ where $f\in M$.

Quillen cohomology is in principle hard to compute, because you need to choose a cofibrant resolution of $M$. However, there is a theorem:
$$
H^t_Q(M;A) \approx H^{t+1}_{HM}(M,\{1\}; A),
$$
where the other side is (relative) "Hochschild-Mitchell" cohomology. Basically, Hochschild-Mitchell cohomology is a Quillen cohomology of $M$, regarded as an object in the category of spaces equipped with a two-sided action by $M$. There is a nice bar complex for computing this.

Anyway, when I work through all this for $M=\{1,f\}$, I think I get the following. The Quillen cohomology $H_Q^t(M,A)$ for $t\geq0$ is equal to the cohomology of a complex:
$$
B \to B \to B \to B \to \cdots.
$$
The group $B=A(f)$. The first differential is $a\mapsto f\cdot a + a\cdot f - a$. The second differential is $a\mapsto f\cdot a-a\cdot f$. Then they alternate.

The two actions $f\cdot, \cdot f$ are commuting idempotents on the abelian group $B$. Thus, you can write elements of $B$ as $2\times 2$-matrices, so that the actions by $f$ amount to left and right multiplication by $\begin{bmatrix} 1&0\\0&0 \end{bmatrix}$. It is then easy to show that
$$
H^t_Q(M, A) =0 \qquad \text{if $t\geq 1$.}
$$
Note that $H^0_Q(M,A)$ is not generally $0$. Thus we ought to conclude that the space $\Map_{s\text{Monoid}}(\widetilde{M}, \Map(X,X))_\gamma$ is connected, with higher homotopy groups $\pi_t= H^0_Q(M,A_t)$.

**Of course, this is not exactly right:** the coefficient system $A_1$ is a *group* object, but may not be an *abelian* group object. So I need to contemplate the non-abelian cohomology
$$
E_2^{1,1} = H^1_Q(M, A_1),\qquad A_1(f) = \pi_1\Map(X,X)_f.
$$
That's more than I want to do right now. In any case, it seems to me the answer to your question comes down to this single group. [Roughly, this group seems to have something to do with "adjusting" a choice of homotopy $\alpha\colon f\sim ff=f$ by a choice of homotopy $\beta\colon f\sim f$; it's something like the orbits of the action $\alpha\mapsto (\beta\beta)\alpha(\beta^{-1})$ on the set $\{\alpha\,|\,\alpha\cdot f=f\cdot \alpha\}$.]

(It's possible, of course, that I've misinterpreted what you mean by "inequivalent" coherentifications. The approach I described presupposes one particular notion of equivalence, but there are others.)

(Or I may have screwed up some other way.)

Higher Algebrashowing that not every homotopy idempotent in spaces splits, even though of course all sequential limits and colimits of spaces exist. $\endgroup$ – Mike Shulman Dec 13 '14 at 0:27