Can a language have $\Sigma^{*}$ as its syntactic monoid? As per the title I was wondering if it's possible for a language $L \subseteq \Sigma^{*}$ to have $\Sigma^{*}$ as its syntactic monoid and if so could one give an example of such a language? I first thought that this was probably an easy question with a simple answer but I haven't been able to make much progress with it, maybe it has a simple answer and I am just overlooking it?

If we look at the syntactic congruence $\sigma_L = \{(w,z) \in \Sigma^{*} \times \Sigma^{*}: (\forall u, v \in \Sigma^{*}) \; uwv \in L \Leftrightarrow uzv \in L\}$ we see that in order to have $w \cong_{\sigma_{L}} z \Leftrightarrow w = z$ we must have that $(\forall w, z \in \Sigma^{*} \; w \not = z) \; (\exists u, v \in \Sigma^{*})\; uwv \in L \Leftrightarrow uzv \notin L$. Which seems like quite a strong condition. It's also clear that such a language couldn't be regular as it is known that a language is regular if and only if it has a finite syntactic monoid.
It seems to work to take $L=\{ww: w\in\Sigma^* \text{and $|ww|$ is a power of $2$}\}$.

First, we can show that if $|u|&gt;|v|$ then they are not equivalent.
Indeed, there is $x$ such that $|uxux|$ is a big power of $2$, say $2^{|u|+|v|}$.
Therefore, $uxux\in L$.

But then the length of $vxux$ cannot be power of $2$, because its difference to $2^{|u|+|v|}$ is only $|u|-|v|$. Therefore, if we take $y=xux$, we have $uy\in L$ and $vy\notin L$.

Now it suffices to show that words of the same length are not equivalent.
Let $u\neq v$ with $|u|=|v|$, and let $x$ such that $|ux|$ is a power of $2$ (and then so is $|uxux|$). Then you can see that $(u,v)\notin\sigma_L$ because $uxux\in L$ and $uxvx\notin L$.