I assume that you mean 'positive definite *symmetric* matrices'. Here's one proof, though it's certainly not the most clean.

A linear automorphism of the space of all symmetric matrices that preserves the cone of positive definite matrices will preserve the closure of that cone, i.e., the *nonnegative* symmetric matrices. The extreme rays of that closed cone are the rank $1$ nonnegative symmetric matrices, i.e., the matrices of the form $s = vv^T$ where $v\in\mathbb{R}^n$ (thought of as columns of height $n$). Now, for any pair of linearly independent vectors $v_1$ and $v_2$ in $\mathbb{R}^n$, the $3$-plane spanned by the matrices of the form $(av_1+bv_2)(av_1+bv_2)^T$ will contain a cone of rank $1$ elements and the automorphism will have to carry that $3$-plane to a $3$-plane that has a cone of rank $1$ elements, and it is easy to see that this implies that this image $3$-plane must be spanned by matrices of the form $(aw_1+bw_2)(aw_1+bw_2)^T$ for some linearly independent pair $w_1$ and $w_2$ in $\mathbb{R}^n$. The upshot of this is that, because all collineations of $\mathbb{RP}^{n-1}$ are projectivizations of linear automorphisms of $\mathbb{R}^n$, there will have to be a linear map $L:\mathbb{R}^n\to\mathbb{R}^n$ such that the automorphism carries $vv^T$ to a positive multiple of $(Lv)(Lv)^T$ for all nonzero $v\in \mathbb{R}^n$. Composing the given automorphism with the automorphism induced by $L^{-1}$, we are reduced to the case of an automorphism of symmetric matrices that carries $vv^T$ to a multiple of $vv^T$ for all nonzero $v\in \mathbb{R}^n$. Of course, since the symmetric matrices have a basis made of rank $1$ elements, it follows that this automorphism must be a (positive) multiple of the identity, so even that can be absorbed into $L$.

coneof the PSD matrices? Clearly, not all automorphisms are given by conjugations, ie maps $X\mapsto B^{-1}XB$, so it's indeed must be $X\mapsto B^\top XB$ – Dima Pasechnik Dec 8 '12 at 16:36