The comments section was getting unwieldy, so I'll answer here. Hopefully this is helpful.

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What I was trying to say is as follows: suppose that $(M,g,X)$ is a steady soliton, i.e.
$$
\mathcal{L}_X(g) = 2 Ric_g.
$$
Then, let $\Phi_t$ denote the flow of $-X$. You may check that $h(t) :=\Phi^*_tg$ is a solution to the Ricci flow. Then, the Ricci curvature of $h(t)$ is 
$$
Ric_{h(t)} = Ric_{\Phi_t^*g} = \Phi_t^*Ric_g.
$$
Thus,
$$
\frac{d}{dt} Ric_{h(t)} = -\mathcal{L}_X Ric_g
$$
Now, relating $\mathcal{L}_X Ric_g$ to $D_X Ric_g$ in the usual way:
\begin{align*}
\mathcal{L}_X Ric(V,W) & = X(Ric(V,W))-Ric([X,V],W)-Ric(V,[X,W])\\
& = D_X Ric(V,W) -Ric(D_XV+[X,V],W)-Ric(V,D_X W +[X,W])\\
& = D_X Ric(V,W) -Ric(D_VX,W)-Ric(V,D_WX)
\end{align*}
and using the formula for $\frac{d}{dt} Ric_{h(t)}$ under the Ricci flow, you may find an equation for $D_X Ric$ (for example, it is the equation found in the paper of Brendle's I linked above; I've matched his sign conventions with $X$). 

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By the way, you can simplify the above argument by using the Uhlenbeck trick (if you work through the details of what I did above, you'll see that there is a good deal of cancellation, which you can exploit by using $D_{\frac{d}{dt}}$ instead of the time derivative above. See, e.g. ch 6 of this book: http://download.springer.com/static/pdf/28/bok%253A978-3-642-16286-2.pdf?auth66=1385658655_8c41399e85b51a3addd04347ad203f0d&ext=.pdf among many other places.