the automorphism of root systems $f(\alpha)=-\alpha$ in the Weyl group I need to find which irreducible root systems have this automorphism in its Weyl group. 
My guess is all. This is what I did:
If $\Delta$ is a basis for the root system  $\Phi$ then $\Delta^- = \{ -\alpha : \alpha \in \Delta\}$ is also a basis. Then by a theorem there exist an element $g \in W$, where $W$ is the Weyl group, such that $g(\Delta)=\Delta^-$. Then since $\delta$ is a basis for $\Phi$ and $f$ and $g$ are the same in $\Delta$ they must be the same automorphism of $\Phi$. 
Is this right? If not what is my mistake and for which irreducible root systems is this true?
 A: I take it that you mean $f = - id$. By the way, your element $g$ also has a name, it's the longest element of the Weyl group (w.r.t. your chosen basis) and is often called $w_0$.
Now, your guess is not quite correct. The mistake is that although on the set level $g(\Delta) = - \Delta$ (i.e. $(f\circ g)(\Delta) = \Delta$), we do not necessarily have $g(\alpha) = -\alpha$ (i.e. $(f\circ g)(\alpha) = \alpha$) for each individual $\alpha \in \Delta$. Actually, e.g. in a root system of type $A_n$ ($n \ge 2$), the following happens: If $\alpha_1, ..., \alpha_n$ is an ordered basis, then $g(\alpha_i) = -\alpha_{n+1-i}$ (i.e. $(f\circ g) (\alpha_i) = \alpha_{n+1-i}$). So $g$ sends the whole basis to its negative, but it does not send each simple root to its own negative, but also reverses the order (i. e. $f\circ g$ stabilises the whole basis as a set, but does not fix it pointwise: it reverses the order).
Your proof shows, however, that $(f\circ g) (\Delta) = \Delta$, which by a theorem means that $f \circ g$ corresponds to an automorphism of the Dynkin diagram. Now in the following cases the only Dynkin diagram automorphism is the identity:
$A_1, B_n, C_n, E_7, E_8, F_4, G_2$.
So in all these cases, $f \circ g = id$ which (because $f^2 = id$) implies $f = g$, so $f \in W$.
The remaining cases are $A_n (n \ge 2), D_n$ and $E_6$. Here one has to start some more subtle arguments (this is done for example in Bourbaki's Lie volume, in the case-by-case discussion at the end of chapter 6; see also Anton Geraschenko's answer here, in which $-w_0$ is exactly our $f\circ g$). It turns out that for $A_{n\ge 2}$, as I wrote above, $f\circ g$ is the non-trivial diagram automorphism, so in particular $f \notin W$. The same is true for type $E_6$.
Finally, type $D_n$ gives rise to another case distinction. Here, $f\circ g$ is the non-trivial diagram automorphism if $n$ is odd, but it's just the identity if $n$ is even. Consequently, $f \notin W$ for type $D_n$ with $n$ odd, but $f\in W$ for type $D_n$ with $n$ even.
Added: Here are two other equivalent criteria for $-id \in W$, which do not rely on case-by-case discussion:
1) All exponents of $W$ are odd. (Reference: Bourbaki, Groupes et algèbres de Lie, ch. V §6 no. 2 corollaire 3 to proposition 3 -- literally the last corollary in that chapter.)
2) The root system contains $l$ mutually orthogonal roots, where $l$ is the rank of the root system. (Reference: R.W. Carter: Conjugacy classes in the Weyl group (Compositio Mathematica 25 (1972), p. 1-59), Lemma 4 on p. 4) (He proves this is necessary. It's easy to see it's sufficient.)
