In the end it was the original question which was answered first.

The answer to Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property? by Ali Enayat shows that there exists a countable ultra-homogeneous structure $M$ with embeddings of $f_i\colon M \to M$, $i = 0,1$, which do not amalgamate inside $M$.

Taking $G = \textrm{Aut}(M)$, $G$ is a Polish group (and moreover homeomorphic to a closed subgroup of $S_\infty$), $f_i \in \hat G$, and there are no $g_i \in \hat G$ such that $g_0 f_0 = g_1 f_1$. This gives the desired counter-example.

(Thank you, Ali!)

In the end it was the original question which was answered first.

The answer to Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property? by Ali Enayat shows that there exists a countable ultra-homogeneous structure $M$ with embeddings of $f_i\colon M \to M$, $i = 0,1$, which do not amalgamate inside $M$.

Taking $G = \textrm{Aut}(M)$, $G$ is a Polish group (and moreover homeomorphic to a closed subgroup of $S_\infty$), $f_i \in \hat G$, and there are no $g_i \in \hat G$ such that $g_0 f_0 = g_1 f_1$. This gives the desired counter-example.

(Thank you, Ali!)

In the end it was the original question which was answered first.

The answer to Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property? by Ali Enayat show that there exists a countable ultra-homogeneous structure $M$ with embeddings of $f_i\colon M \to M$, $i = 0,1$, which do not amalgamate inside $M$.

Taking $G = Aut(M)$, $G$ is a Polish group (and moreover homeomorphic to a closed subgroup of $S_\infty$), $f_i \in \hat G$, and there are no $g_i \in \hat G$ such that $g_0 f_0 = g_1 f_1$. This gives the desired counter-example.

(Thank you, Ali!)

In the end it was the original question which was answered first.

The answer to Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property? by Ali Enayat shows that there exists a countable ultra-homogeneous structure $M$ with embeddings of $f_i\colon M \to M$, $i = 0,1$, which do not amalgamate inside $M$.

Taking $G = \textrm{Aut}(M)$, $G$ is a Polish group (and moreover homeomorphic to a closed subgroup of $S_\infty$), $f_i \in \hat G$, and there are no $g_i \in \hat G$ such that $g_0 f_0 = g_1 f_1$. This gives the desired counter-example.

(Thank you, Ali!)