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 Shi (Deforming the metric on complete Riemannian manifold, JDG 1989) 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 Chen-Zhu (Uniqueness of the Ricci Flow on Complete Noncompact Manifolds) 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 Kotschwar (Backwards Uniqueness of Ricci Flow). 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 Chen (Strong Uniqueness of the Ricci flow, JDG 2009), 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.

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 Shi (Deforming the metric on complete Riemannian manifold, JDG 1989) 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 Chen-Zhu (Uniqueness of the Ricci Flow on Complete Noncompact Manifolds) 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 Kotschwar (Backwards Uniqueness of Ricci Flow). 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 Chen (Strong Uniqueness of the Ricci flow, JDG 2009), 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.