I assume you are talking about representations of $S_n$ over $\mathbb{C}$. In this case, $\mathbb{C}S_n$ is left semi-simple by Maschke's Theorem, so every representation of $S_n$ is a direct sum of irreducible representations.

Next is Schur's Lemma: Let $V$ and $W$ be irreducible representations of $S_n$. Then any homormorphism $\phi:V\to W$ is either 0 or an isomorphism.

proof: $\phi(V)\subset W$ is an $S_n$ submodule. QED

Corollary: The map $\phi$ above is scalar multiplication by a constant.

proof: $V$ is a $\mathbb{C}$-vector space, hence $\phi$ has an eigenvector $v\in V$ with $\phi(v) = cv$ for some $c\in \mathbb{C}$. Then $v\in\mathrm{ker}(\phi-cId_V)\neq 0$ so $\phi=cId_V$ by Schur's lemma. QED

Now, let $V$ be irreducible and $W$ be any submodule. Write $W=\bigoplus_i V_i^{\oplus m_i}$ for non-isomorphic $V_i$. By the previous two statements, any nonzero map $\phi:V\to W$ must factor as $$\phi:V\cong V_i\hookrightarrow W$$ for some $i$. This proves Q1. Oh, and Q2 too (after Abdelmalek Abdesselam's correction).

I assume you are talking about representations of $S_n$ over $\mathbb{C}$. In this case, $\mathbb{C}S_n$ is left semi-simple by Maschke's Theorem, so every representation of $S_n$ is a direct sum of irreducible representations.

Next is Schur's Lemma: Let $V$ and $W$ be irreducible representations of $S_n$. Then any homormorphism $\phi:V\to W$ is either 0 or an isomorphism.

proof: $\phi(V)\subset W$ and $\ker(\phi)\subset V$ are $S_n$ submodules. QED

Corollary: The map $\phi$ above is scalar multiplication by a constant.

proof: $V$ is a $\mathbb{C}$-vector space, hence $\phi$ has an eigenvector $v\in V$ with $\phi(v) = cv$ for some $c\in \mathbb{C}$. Then $v\in\mathrm{ker}(\phi-cId_V)\neq 0$ so $\phi=cId_V$ by Schur's lemma. QED

Now, let $V$ be irreducible and $W$ be any submodule. Write $W=\bigoplus_i V_i^{\oplus m_i}$ for non-isomorphic $V_i$. By the previous two statements, any nonzero map $\phi:V\to W$ must factor as $$\phi:V\cong V_i\hookrightarrow W$$ for some $i$. This proves Q1. Oh, and Q2 too (after Abdelmalek Abdesselam's correction).