Suppose, without loss of generality, that $\Omega = \{1,2,3\}$. Recall that ${\rm E}[X|Y]$ stands for ${\rm E}[X|\sigma(Y)]$, where $\sigma(Y)$ is the $\sigma$-algebra generated by $Y$. Let ${\rm P}$ be the uniform probability measure on $\Omega$, that is, ${\rm P}(\{\omega\})=1/3$, $\forall \omega \in \Omega$. Now, define random variables $X$, $Y$, and $Z$ as follows: $$ X(\omega)=\omega,\;\; \forall \omega \in \Omega , $$ $$ Y(1)=1, Y(\omega)=2.5, \omega \in \{2,3\}, $$ and $$ Z(\omega)=1.5, \omega \in \{1,2\}, Z(3)=3. $$ Then, $$ {\rm E}[X|Y] = Y $$ and $$ {\rm E}[X|Z] = Z. $$ It thus suffices to show that $$ {\rm E}[Y|Z] \ne {\rm E}[Z|Y]. $$ Indeed, ${\rm E}[Y|Z]$ takes the values $1.75$ and $2.5$ with probabilities $2/3$ and $1/3$, respectively, whereas ${\rm E}[Z|Y]$ takes the values $2.25$ and $1.5$ with probabilities $2/3$ and $1/3$, respectively. So, these random variables are never equal.