No. Let $G$ be a nontrivial group under the operation $\bullet$ and let $f:G\to G$ be a permutation of its elements (which we don't regard as a homomorphism). Define $\circ$ on $G$ by $f(a\circ b)=f(a)\bullet f(b)$.
Then $(G,\circ)\cong(G,\bullet)$ but $\circ,\bullet$ needn't be the same operation. Essentially this means we can make the underlying set of $G$ into the group $G$ in many different ways by relabelling its elements.