If $K=K^2$ then every automorphism of $\mbox{Aut}_K V$, where $\dim V< \infty$, is the square of some endomorphism. I have to show the following: 

Let $K$ be a field such that $\mbox{char } K \neq 2$ and each element of $K$ is a square (i.e. $K^2=K$) and let $V$ be a finite-dimensional vector spaces over $K$. Then, for every automorphism $\tau \in \mbox{Aut}_K V$ there exists an endomorphism $\rho \in \mbox{End}_K V$ such that $\tau = \rho^2$.

I have proved (according to the hint given in the problem) that if $\sigma$ is a nilpotent endomorphism, then there exists an endomorphism $\rho$ such that $\rho^2=1_V+\sigma$. 
So, I guess (although I am not sure) that under our assumptions one could show the automorphism $\tau$ can be represented as $\tau=1_V+\sigma$, where $\sigma$ is nilpotent. I'll be grateful for your help.
 A: If my answer to the question Reducibility of $P(X^2)$
appears to be right, then I think the statement is false.
1 — Counter example
Consider $A$ the companion matrix of a polynomial $P(x)$, and let $B$ be a square root of $A$. It is clear — take a triangulation in the algebraic closure — that $\chi_B(x) \chi_B(-x) = \chi_A(x^2)$ and that $\chi_A(x) = P(x)$.
This implies that $P(x^2)$ is reducible over the coefficient field of $B$. If one take $P = x^5 + 20x - 16$, I think I have proved that $P(x^2)$ is irreducible over the quadratic closure of $\Bbb Q$, which implies that $A$ has no square root over this field.
2 — Complement
Let $K$ be a field, with characteristic not two, and $A$ a square matrix with coefficients in $K$.
It is not easy to see whether of not $A$ admit a square root.
We can assume that $\chi_A$ is the power of a irreducible polynomial. Indeed, $A$ stabilize its eigen spaces associated to each irreducible factor, and so does every matrix $B$ which commutes with $A$, which is the case if $B^2 = A$.
There nilpotent case — corresponding to $\chi_A = x^n$ — is particular, there is some combinatorial condition on the size of the nilpotents Jordan blocks.
Let consider the non-singular case. We can assume that $A$ is diagonalizable. Indeed, we can always write $A$ as $D+N$, with $D$ diagonalizable and $N$ nilpotent, both with coefficients in $K$, and with $DN = ND$. The matrix $A$ has a square root if and only if $D$ has a square root.
So we are reduced to the case of a matrix with blocks along the diagonal all equal to the companion matrix $C_P$, where $\chi_A = P^d$. This matrix has a square root if and only if one of the following holds :


*

*$n$ is even ;

*The decomposition field of $\chi_A$ contains the roots of $\chi_A(x^2)$.

