Let $M$ denote the $F$-algebra $\hom(V, V)$. An $F$-algebra homomorphism $M \to \hom(V, V)$ is tantamount to an $M$-module structure on $V$ in the category of $F$-vector spaces. But the category of $M$-modules is equivalent to $Vect_F$ (there is a Morita equivalence between the categories given by the functor $- \otimes_F V: Vect_F \to Mod_M$). That is to say, every $M$-module is isomorphic to a direct sum of copies of $V$, and in particular, any module structure on $V$ is isomorphic to the standard $M$-module structure on $V$.
So, if $\mu': M \otimes_F V \to V$ is any module structure, and $\mu: M \otimes_F V \to V$ denotes the standard structure, there is an isomorphism $g: V \to V$ such that $g \circ \mu' = \mu \circ (M \otimes_F g)$. Putting all this together, any $F$-algebra homomorphism $M \to \hom(V, V)$ is obtained by conjugating the standard homomorphism (the identity) by an isomorphism $g: V \to V$.
Edit: If this proof seems too categorical, there is a proof in slightly alternative language given on page 401 of Bilinear algebra: An Introduction to the algebraic theory of quadratic forms by Szymiczek (Google books here.