$E$ a $\mathbb{C}$-vector space of dimension $n>2$
Let $f : E \rightarrow E$ an endomorphisme which commutes with all automorphisms of $E$.
Show that $f$ is a homothety
Let $\lambda$ an eigenvalue of $f$ and $x$ is an eigenvector. For any automorphism $g$ we have:
$f(g(x))=g(f(x))=\lambda g(x)$. So $g(x)$ is an eigenvector.
We complete $(x)$ in a basis $(x, e_2, \dots, e_n)$ of $E$
By choosing $n-1$ automorphisms of $g_i$ such that $g_i(x)=e_i$ for $2 \leq i \leq n$, we get that $f$ is a homothety (on the basis).
Now let $f : E \rightarrow E$ an application, not assumed to be linear. Show that that $f$ is always a homothety.
Here I do not see how can I proceed.
If $f$ is just an application:
(1) let $u \in E$, $f$ commutes with the orthogonal symmetry w.r.t $\mathbb{C} \cdot u$, that we note $s$. Thus, we have $f(s(u))=s(f(u))$ but $s(u)=u$, and $f(u)=s(f(u))$. That shows that $f(u)$ is stabilized by $s$, thus $f(u) \in \mathbb{C} \cdot u$, id est $f(u) = \lambda_u u$.
(2) Now, let show that this coefficient is the same for all vectors : let $u,v \in E - \{ 0 \}$, and we note $r$ any linear automorphism that send $u$ to $v$ (for example, a planar rotation composed with a homothety will do). Then you have $f(r(u))=rf(u)$, but $r(u)=v$ thus $f(v)=rf(u)$. Now, according to (1), $f(u)=\lambda_u u$, $f(v) = \lambda_v v$ then $$\lambda_v v = r(\lambda_u u) = \lambda_u r(u) = \lambda_u v$$. Finally, $\lambda_u=\lambda_v$ and for any $v \in E$, $f(v)=\lambda_u v = \lambda v$