Lemma:

Let $f:W \to V$ be an homogeneous algebraic map (i.e. $f(\alpha v)=\alpha^k f(v)$ for some $k$). Then the image of $f$ is closed sub-variety of $V$.

Proof: let $\bar f: \mathbb P(W) \to \mathbb P(V)$ be the corresponding map of protective spaces. Since $\mathbb P(W) $ is complete, the image of $\bar f$ is a closed sub-variety of $\mathbb P(V)$. So it is given be a collection of homogeneous polynomials on $V$. The same polynomials are the one that defines the image of $f$.

The map $m \mapsto mv$ is clearly homogeneous map of degree 1. so the lemma implies that the image is a closed sub-variety of $V$.

Did I got something wrong?

As explained in the comments, the following Lemma is wrong:

Let $f:W \to V$ be an homogeneous algebraic map (i.e. $f(\alpha v)=\alpha^k f(v)$ for some $k$). Then the image of $f$ is closed sub-variety of $V$.

Proof: let $\bar f: \mathbb P(W) \to \mathbb P(V)$ be the corresponding map of protective spaces. Since $\mathbb P(W) $ is complete, the image of $\bar f$ is a closed sub-variety of $\mathbb P(V)$. So it is given be a collection of homogeneous polynomials on $V$. The same polynomials are the one that defines the image of $f$.

The rest is probably correct now (I had to correct it too), but probably useless:

The map $m \mapsto mv$ is homogeneous map (since the representation $V$ is irreducible).