I was going to post this as a comment, but it got too long. I'm not exactly sure how to answer the question as written, but the following might be enlightening: 

A crucial piece of information for your Ricci flow is that you start with bounded curvature, i.e. 
$$
\sup_M |Riem|_{g(0)} < \infty.
$$
Then, according to <a href="http://www.intlpress.com/JDG/archive/1989/30-1-223.pdf">Shi (Deforming the metric on complete Riemannian manifold, JDG 1989)</a> there _exists_ a Ricci flow $g(t)$ with bounded curvature, $g(t)$, defined for $t \in [0,T)$ where for $t \in [0,T)$
$$
\sup_M |Riem|_{g(t)} < \infty.
$$
Shi makes no claims about uniqueness.

Later, it was proved by <a href="http://arxiv.org/abs/math/0505447">Chen-Zhu (Uniqueness of the Ricci Flow on Complete Noncompact Manifolds)</a> that if $g_1(t), g_2(t)$ are solutions to Ricci flow for $t \in [0,T)$ and they _both_ have bounded curvature in $[0,T)$, _and_ $g_1(0) = g_2(0)$, then $g_1(t) = g_2(t)$ 

In particular, a corollary of Chen-Zhu's result is:
> If $\phi\in Isom(M,g(0))$ is an isometry of $(M,g(0))$ where $g(0)$ has bounded curvature, then _the_ Shi solution to Ricci flow with initial data $g(0)$ still has $\phi \in Isom(M,g(t))$.

> A "fancy" way of saying this is $Isom(M,g(0)) \subseteq Isom(M,g(t))$.

To prove this, if $\phi \in Isom(M,g(0))$, let $g(t)$ be the Shi solution to Ricci flow starting at $g(0)$. Clearly $\phi^*g(t)$ is still a solution to Ricci flow, but $\phi^*g(0) = g(0)$ by assumption that it is an isometry of $g(0)$. So $\phi^*g(t)$ is a solution to Ricci flow (obviously having bounded curvature) with initial data $g(0)$, so it must be $g(t)$!

I assume that you were originally referring to <a href="http://arxiv.org/pdf/0906.4920v1.pdf">Kotschwar (Backwards Uniqueness of Ricci Flow)</a>. This result says the opposite of Chen-Zhu, and is not really what you want here. In particular, it says that if $g_1(t), g_2(t)$ are both solutions to Ricci flow on $t\in[0,T]$ with uniformly bounded curvature ($T<\infty$), if $g_1(T) = g_2(T)$ then $g_1(t) = g_2(t)$ for all $t \in [0,T]$. In particular, this complements the Chen-Zhu corollary:
> Under the assumptions of uniformly bounded curvature, $Isom(M,g(t)) \subseteq Isom(M,g(0))$.


Finally, I'll remark on the assumption of bounded curvature. In <a href="http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jdg/1246888488">Chen (Strong Uniqueness of the Ricci flow, JDG 2009)</a>, it is shown that one does not always need to assume bounded curvature in dimension $3$. Instead, Chen shows:
> If $(M,g(0))$ is complete, has bounded, nonnegative sectional curvature, $0 \leq Rm \leq K$, then any two $g_1(t),g_2(t)$ Ricci flows (which are complete for all $t\in[0,T)$) starting from $g(0)$ must agree.



---

None of this exactly answers your question as it stands, but I don't think that being simply connected makes things any easier (but I could be wrong). On the other hand, as long as your initial manifold has bounded curvature, by Shi-Chen-Zhu, there is a unique (short time) solution to Ricci flow with bounded curvature.