A First Course in Graph Theory by Gary Chartrand and Pink Zhang (2012) defines an isomorphism as so:
Two labeled graphs $G$ and $H$ are isomorphic (have the same structure) if there exists a one-to-one correspondence $\phi$ from $V(G)$ to $V(H)$ such that $uv\in E(G)$ if and only if $\phi(u)\phi(v)\in E(H)$.
Symbolically, I would represent this as:
$\left(\exists \phi: V(G)\to V(H) \text{ such that } uv\in E(G)\iff \phi(U)\phi(v)\in E(H)\right)\implies G\cong H$
Is this definition an if and only if? That is, can I say this:
$G\cong H\iff\left(\exists \phi: V(G)\to V(H) \text{ such that } uv\in E(G)\iff \phi(U)\phi(v)\in E(H)\right)$
Wolfram Alpha and various Google results do not use an if and only if statement.
Yes, this is implicitly an "if and only if" because it is a definition - the description given is a characteristic property of what it means for two graphs to be isomorphic.
In one direction, by definition, any two graphs satisfying this property are called isomorphic. But also by the definition of isomorphism it means that if two graphs do not satisfy this property, then they are not considered isomorphic, which proves the converse.
In mathematical literature, the usage of "if" in a definition (X is called Y if Z) is always "iff".