Let $\phi: V \to W$ be a non-zero $G$-homomorphism. We must show this leads to an isomorphism, contradicting $V \not\cong W$.

Step 1: The kernel $\ker(\phi) = \{v \in V : \phi(v) = 0\}$ is a subrepresentation of $V$.

Proof: If $v \in \ker(\phi)$ and $g \in G$, then: $\phi(g \cdot v) = g \cdot \phi(v) = g \cdot 0 = 0$ so $g \cdot v \in \ker(\phi)$. Thus $\ker(\phi)$ is $G$-invariant.

Step 2: Since $V$ is irreducible and $\ker(\phi) \neq V$ (as $\phi \neq 0$), we have $\ker(\phi) = \{0\}$, so $\phi$ is injective.

Step 3: The image $\text{im}(\phi) = \{\phi(v) : v \in V\}$ is a subrepresentation of $W$.

Proof: If $w = \phi(v) \in \text{im}(\phi)$ and $g \in G$, then: $g \cdot w = g \cdot \phi(v) = \phi(g \cdot v) \in \text{im}(\phi)$

Step 4: Since $W$ is irreducible and $\text{im}(\phi) \neq \{0\}$ (as $\phi \neq 0$), we have $\text{im}(\phi) = W$, so $\phi$ is surjective.

Conclusion: $\phi$ is bijective, hence an isomorphism $V \cong W$. Therefore, if $V \not\cong W$, no non-zero homomorphism can exist.

Let $\phi: V \to V$ be a $G$-homomorphism. We use the assumption that $\mathbb{F}$ is algebraically closed.

Step 1: Since $\mathbb{F}$ is algebraically closed, the characteristic polynomial of $\phi$ has a root $\lambda \in \mathbb{F}$. This means $\lambda$ is an eigenvalue of $\phi$.

Step 2: Consider the operator $\psi = \phi - \lambda \cdot \text{id}_V$. This is also a $G$-homomorphism: $\psi(g \cdot v) = \phi(g \cdot v) - \lambda(g \cdot v) = g \cdot \phi(v) - g \cdot (\lambda v) = g \cdot \psi(v)$

Step 3: The kernel $\ker(\psi)$ is the $\lambda$-eigenspace of $\phi$, which is non-empty by Step 1. Moreover, $\ker(\psi)$ is a subrepresentation of $V$ (by the argument in Part 1).

Step 4: Since $V$ is irreducible and $\ker(\psi) \neq \{0\}$, we must have $\ker(\psi) = V$. This means $\psi = 0$, i.e., $\phi = \lambda \cdot \text{id}_V$.

Conclusion: Every endomorphism of $V$ is a scalar multiple of the identity.