Isomorphism notion between sets of triples
Your problem reduces to graph isomorphism, so is not $\mathsf{NP}$-complete unless the polynomial hierarchy collapses. 

In fact, you've essentially already given the reduction in your question: given a 3-uniform (=every hyperedge has size 3) hypergraph $G$, construct from it the graph $f(G)$ which has the same vertex set, and such that $\{u,v\} \in E(f(G))$ iff there exists a $z \in V$ such that $\{u,v,z\} \in E(G)$. (When $G$ is a hypergraph, I use $E(G)$ to mean the set of hyperedges.) Then $G_1$ and $G_2$ are isomorphic-in-your-sense iff $f(G_1) \cong f(G_2)$ as graphs. It is not hard to check that this reduction only takes polynomial time (note that $|E(f(G))| \leq |V(G)| |E(G)|$). 

It is also probably worth mentioning that there is already a well-established notion of hypergraph isomorphism that is different from your notion. If we consider a hypergraph $G$ as a vertex set $V$ together with a collection of hyperedges $E \subseteq P(V)$ (each hyperedge is a subset of $V$), then two such hypergraphs are isomorphic if there is a bijection $\pi\colon V_1 \to V_2$ such that $E_2 = \pi(E_1)$. (Here I am abusing notation twice: by "$\pi(E_1)$" I mean $\{\pi(e) : e \in E_1\}$ and by $\pi(e)$ for $e \subseteq V$ I mean $\{\pi(v) : v \in e\}$.) Hypergraph isomorphism in this sense is Karp-equivalent to graph isomorphism, but by a slightly less immediate reduction than from your problem. However, fast (in theory) algorithms for hypergraph isomorphism are often more difficult than those for graph isomorphism (see, e.g. Babai and Codenotti, FOCS 2008).