you didn't menstion but i guess your base field is well, this is what we can say: where is the identity map. the reason is this:

let and then for all and we'll have that means is a module homomorphism.

also, since is algebraically closed, has an eigenvalue, say thus is a module homomorphism which is not an isomorphism because it has a non-zero kernel.

thus, since is irreducible, we must have by Schur's lemma.