Here's my favorite way to answer your question.  Hopefully the answer to Robert Bryant's question is "yes".

Let $A$ be the ring of octonions (the "nonsplit" octonions over $R$); it comes with an involution $\alpha \mapsto \bar \alpha$, from which there is a trace $Tr(\alpha) = \alpha + \bar \alpha$ and a norm $N(\alpha) = \alpha \cdot \bar \alpha$.  From this, we get a trilinear form $T: A \otimes A \otimes A \rightarrow R$ given by 
$$T(\alpha, \beta, \gamma) = Tr( (\alpha \beta) \gamma) = Tr(\alpha (\beta \gamma)).$$
(Multiplication is nonassociative, but the traces work out to the same result.)

The group $Spin(8)$ can be constructed as the group of triples $(g_1, g_2, g_3) \in SO(A,N)^3$ of "rotation matrices" with respect to the norm quadratic form, such that for all $(\alpha, \beta, \gamma) \in A^3$, 
$$T(g_1 \alpha, g_2 \beta, g_3 \gamma) = T(\alpha, \beta, \gamma).$$

The full group of outer automorphisms is now clear -- just permute $(g_1, g_2, g_3)$ via the symmetric group $S_3$.  

The Jordan algebra is the exceptional Jordan algebra of 3x3 Hermitian matrices with octonion entries:
$$J = \{
\left( \begin{array}{ccc} 
a & \alpha & \bar \beta \\ 
\bar \alpha & b & \gamma \\ 
\beta & \bar \gamma & c \\
\end{array} \right) : \alpha, \beta, \gamma \in A, a,b,c \in R \}.$$

(Forgive me -- I can't get my 3x3 matrix or my braces to work here!)

The group $Spin(8)$ acts on the triple of octonions $(\alpha, \beta, \gamma)$ via the natural representation from above.  It acts trivially on the real numbers $a,b,c$, and this gives an action of $Spin(8)$ on the exceptional Jordan algebra.  The outer automorphism group $S_3$ acts by permuting $(a,b,c)$ and $(\alpha, \beta, \gamma)$ simultaneously.  Together, these give an action of $S_3 \ltimes Spin(8)$ on the exceptional Jordan algebra.