Regular language where syntactic right congruence and syntactic congruence differ
Definition: An equivalence relation $\sim$ on $\Sigma^*$ is a right (left) congruence if $x\sim y$ implies $xv\sim yv$ (respectively $vx\sim vy$) for every $x, y, v \in\Sigma^*$.
Definition: Given a language $L$ over an alphabet $\Sigma$, the canonical right congruence $\sim_L$ is defined by $x\sim_L y$ if and only if for all $z\in\Sigma^*$ we have $xz\in L\iff yz\in L$. Similarly, for the canonical left congruence $x{}_{L}\sim y$ if for all $z\in\Sigma^*$ we have that $zx\in L\iff zy\in L$.
Example: Consider the (regular) language $L=\{ab,abb\}$ over $\Sigma^*$. For this language, the relations $\sim_L$ and ${}_L\sim$ are not the same.
In fact, according to ${}_L\sim$ the words $x=ab$ and $y=abb$ are in the same class. This is because if we prepend any $z\in\Sigma^*$ to $x$ and $y$, both word $zx$ and $zy$ are in $L$ or not simultaneously. In fact, for $z=\epsilon$ both $zx=\epsilon ab=ab$ and $zy=\epsilon abb=abb$ are in $L$. However, when $z\neq \epsilon$, then both $zx$ and $zy$ are not in $L$. Therefore $ab{}_L\sim abb$.
Now, when we consider the relation $\sim_L$, we have that $ab\not\sim_L abb$. In fact, if we take $z=b$ we get that $xz=abb\in L$ but $yz=abbb\not\in L$.
So, the regular language $L$ is an example for which its left and right canonical congruences are not the same relation.