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 <a href="http://books.google.com/books?id=CcM8_iiGPxAC&pg=PA401&lpg=PA401&dq=automorphisms+of+a+matrix+algebra&source=bl&ots=Ba5Y2F3AbT&sig=qX8ef4eQVkLoFN-_5CtHGX4s8h8&hl=en&ei=7hm5TJyVOs6s8AauyfH-Dg&sa=X&oi=book_result&ct=result&resnum=7&ved=0CDQQ6AEwBjgK#v=onepage&q=automorphisms%20of%20a%20matrix%20algebra&f=false">here</a>.