= Solution
Take
$$
I_0=[x_0,x_2],\qquad I_1=[x_2,x_1],\qquad I_2=[x_1,x_3].
$$
Now
$$
I_0\longrightarrow I_2,\qquad
I_1\longrightarrow I_1,I_2,\qquad
I_2\longrightarrow I_0,
$$
so
$$
A=\begin{pmatrix}0&0&1\\0&1&1\\1&0&0\end{pmatrix}.
$$
The recurrent pieces of this <directed covering graph of an interval map>[covering graph] are the two-cycle $I_0\leftrightarrow I_2$ and the loop at $I_1$; neither contains two competing closed routes. Also
$$
\operatorname{tr}(A^3)=\operatorname{tr}A=1,
$$
so this ordering forces no 3-cycle.
The bound is attained by the <piecewise linear function>
$$
F(x)=
\begin{cases}
x+\dfrac23,&0\leq x\leq\dfrac13,\\[3pt]
\dfrac53-2x,&\dfrac13\leq x\leq\dfrac23,\\[3pt]
1-x,&\dfrac23\leq x\leq1.
\end{cases}
$$
Indeed,
$$
0\mapsto\frac23\mapsto\frac13\mapsto1\mapsto0
$$
is a 4-cycle with the required spatial order. On $I_0\cup I_2$, the fourth iterate is the <identity function>. Within $I_1$, the fixed point is $5/9$ and
$$
F(x)-\frac59=-2\left(x-\frac59\right),
$$
so every other point eventually leaves $I_1$ and enters $I_0\cup I_2$. The map therefore has no horseshoe in any iterate and is not <Glendinning chaos>[chaotic]. The minimum number of distinct 3-cycles is consequently $\boxed0$.
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