Am I correct in thinking that the category pGrp, whose objects are pairs $(G,g)$ where $G$ is a group, $g \in G$ and 

$$\hom\left((G,g),(H,h) \right) = \{ \varphi: G \rightarrow H \hspace{1mm} \big\vert \hspace{1mm} \varphi(g) = h \},$$

is neither isomorphic to nor equivalent to the regular category Grp? 

The counterexample I was thinking of was that $\hom\left((\mathbb{Z},2), (\mathbb{Z},3)\right) = \{ \} $, but that, under the forgetful functor $F$ which essentially forgets the points and leaves the groups and homomorphisms intact, the corresponding hom-set would be $\hom \left( \mathbb{Z},\mathbb{Z} \right) $, an infinite set (since we can have $\varphi(x) = nx $ for any $ n \in \mathbb{Z}$), so $F$ would not be full. 

I suppose technically this wouldn't constitute a proof (it would only be a proof that $F$ is not an equivalence), since there are other functors from pGrp to Grp, but is my line of thinking generally correct?

Thanks!