Important: the following answer is assuming $|R|<\infty$, otherwise this isn't true in general.
Consider the ring homomorphism $\varphi: R\to R$. Then, if you want to prove it's an isomorphism, you just need to prove it's injective or it's surjective (you don't need to prove both).
Why is this true? Because, since $\varphi$ goes from $R$ to $R$ ($R$ goes to himself), the cardinality of the origin set and the destiny set are the same (because they are the same set). So:
If $\varphi$ is injective, every different $x\in R$ gives a different $\varphi (x)$, in fact $|R|$ different images, but the cardinality of the destiny set is $R$ so it's also surjective, hence a bijection.
If $\varphi$ is surjective, then every $y\in R$ has some preimage $x$, so in order for that to be true you need $|R|$ different preimages, the cardinality of the origin set, so then $\varphi$ is also injective, hence a bijection.
So, when facing an homomorphism that goes from one finite set to himself, you just need to prove one of the two conditions to prove it's a bijection (since the other one will be a direct consequence of the first).