Given any elliptic curve over $\mathbb{Q}$ of conductor $N$, by modularity of elliptic curves,
there exists a surjective morphism from $X\_0(N)$ $\rightarrow$ $E$.There may be several such 'N' and morphisms from $X\_0(N)$ $\rightarrow$ $E$. One such $N$ is the conductor of $E$.


If there exists an elliptic curve $E$ and a morphism $f$: $X\_0(p)$ $\rightarrow$ $E$ for a fixed prime $p$, then "will it imply that the conductor of $E$ is $p$".



Conductor of an elliptic curve
Yes. If there is a surjection $X\_0(N) \to E$ then the conductor of $E$ must divide $N$. Since there is no elliptic curve of conductor 1, an elliptic curve uniformized by $X\_0(p)$ for $p$ prime must have conductor $p$.


