Yes, the only isomorphisms in the free category of any graph are the identity morphisms. This follows from the fact that the morphisms in \\(\bf{Free}(G)\\) do not obey any equations (other than those required in the definition of a category, i.e., the left and right unit laws).

Suppose, by way of contradiction, that \\(f:A\to B\\) is a non-identity isomorphism in \\(\bf{Free}(G)(A,B)\\). Then there is a morphism \\(g:B\to A\\) such that \\(f\circ g=\text{id}_A\\) and \\(g\circ f=\text{id}_B\\). However, since \\(f\\) is not an identity morphism and the only equations among elements of \\(\bf{Free}(G)(A,B)\\) are the left and right unit laws, no such morphism \\(g\\) exists. In particular, if \\(A\neq B\\), then there are no equations at all between elements of \\(\bf{Free}(G)(A,B)\\). Even if \\(A=B\\), the only option is \\(g=\text{id}_A\\), which yields \\(f\circ\text{id}_A=\text{id}_A\\), so that \\(f=\text{id}_A\\), which contradicts the assumption that \\(f\\) is not the identity morphism.

Suppose, by way of contradiction, that \\(f:A\to B\\) is a non-identity isomorphism in \\(\bf{Free}(G)(A,B)\\). Then there is a morphism \\(g:B\to A\\) such that \\(f\circ g=\text{id}_A\\) and \\(g\circ f=\text{id}_B\\). However, since \\(f\\) is not an identity morphism and the only equations among elements of \\(\bf{Free}(G)(A,B)\\) are the left and right unit laws, no such morphism \\(g\\) exists. In particular, if \\(A\neq B\\), then there are no equations at all between elements of \\(\bf{Free}(G)(A,B)\\). Even if \\(A=B\\), the only option is \\(g=\text{id}_A\\), which yields \\(f\circ\text{id}_A=\text{id}_A\\), so that \\(f=\text{id}_A\\), which contradicts the assumption that \\(f\\) is not the identity morphism.