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.