Is is possible to have a pair commuting observables only in a single direction? In quantum mechanics, for two observables to be compatible, successive measurements of the observables, say $A$ and $B$, should yield the same result as earlier, i.e if we do the measurements with the order $A \to B \to A$, the result from the first $A$ and the last $A$ should be the same, and similarly, for $B \to A \to B$.
However, is it possible for two observables to have the following relation;
If we measure $A \to B \to A$, the first and the last measurement of $A$ yield the same result, but if we measure $B\to A \to B$, the first and the last measurement of $B$ yields different measurements (in general).
 A: I'll start with a couple of caveats and then try to answer the question.


*

*First caveat: Consecutive measurements of the same observable $A$ don't necessarily yield the same outcome, not even if no other observable $B$ is measured in between, because the system evolves according to the Schrödinger equation between the two $A$-measurements. (For example, the more precisely we try to measure a particle's position, the more rapidly it loses that precise precision after the measurement is done, because $\Delta x\,\Delta p\geq \hbar/2$.) With some idealization, though, we can say that two immediately-consecutive measurements of $A$ would yield the same outcome. The following answer uses this projective measurement idealization. 

*One more caveat: Implicit in the projective-measurement idealization is the assumption that the observables in question are discrete, because only then do they have normalizable eigenstates. (This is implicit in the idea of a sharply-defined "outcome.") For that reason, the following answer considers only discrete observables. Since any continuous observable can be arbitrarily-well approximated by a discrete one, this should be sufficient for all practical purposes.
With these idealizations/caveats, here is the answer to the question:


*

*Suppose that observables $A$ and $B$ have the property that if we measure $A$-then-$B$-then-$A$ with no delay, the outcomes of the first and last $A$-measurement are always the same.

*This implies that if we measure $B$-then-$A$-then-$B$ with no delay, then the outcomes of the first and last $B$-measurement are always the same. In the OP's words, two observables cannot commute in only a single direction.
Proof: The observables' eigenvalues are not relevant to this question, so we can think of the discrete observable $A$ as a set 
$$
A=\{A_1,A_2,...\}
\tag{1}
$$ 
of mutually commuting projection operators $A_j$ satisfying
$$
  \sum_j A_j=1
\hskip2cm
   A_j A_k=0\text{ if }j\neq k
\tag{2}
$$
and similarly for the other observable $B=\{B_1,B_2,...\}$. Now, suppose that that whenever we measure $A$ then $B$ then $A$, the outcomes of the two $A$-measurements are always the same. If we start with any state-vector $|\psi\rangle$ and apply an $A$-measurement followed by a $B$-measurement, with outcomes $A_j$ and $B_k$, respectively, then the resulting state-vector is $B_kA_j|\psi\rangle$. We have assumed that the outcome of a subsequent $A$-measurement can only be $A_j$, which implies 
$$
A_mB_k A_j|\psi\rangle=0
\hskip1cm
\text{ for all $k,\psi$ whenever } m\neq j. 
\tag{3}
$$
This holds for all $\psi$, so
$$
A_mB_k A_j=0
\hskip1cm
\text{ for all $k$ whenever } m\neq j.
\tag{4}
$$
Equation (2) implies
$$
\sum_m A_mB_k A_j = B_k A_j
\hskip1cm
\text{for all }k,j
\tag{5a}
$$
and equation (4) implies
$$
\sum_m A_mB_k A_j = A_j B_k A_j
\hskip1cm
\text{for all }k,j.
\tag{5b}
$$
Combine equations (5a) and (5b) to get
$$
    B_kA_j = A_jB_kA_j
\hskip1cm
\text{for all }k,j.
\tag{6}
$$
Taking the adjoint of both sides of (6) gives
$$
    A_jB_k = A_jB_kA_j
\hskip1cm
\text{for all }k,j
\tag{7}
$$
because projection operators are self-adjoint and because the adjoint reverses the order of multiplication. Combine (6) and (7) to get $A_jB_k=B_k A_j$ for all $j,k$. This impiles that the observables $A$ and $B$ commute with each other, which completes the proof.
