[proofplan] Pull back the given Ricci flow by the initial isometry $\phi$ and call the resulting family $h(t)=\phi^*g(t)$. The Ricci tensor is natural under pullback by diffeomorphisms, so $h(t)$ satisfies the same Ricci flow equation as $g(t)$. Since $\phi$ is an isometry of the initial metric, both flows have the same initial data. Hamilton uniqueness for Ricci flow on closed manifolds then forces the two solutions to coincide for every time in their common interval of existence. [/proofplan]

[step:Pull back the evolving metric by the initial isometry] Let $\Gamma(S^2T^*M)$ denote the space of smooth sections of the bundle $S^2T^*M$ of symmetric covariant $2$-tensors on $M$. Define the map \begin{align*} h: [0,T) \to \Gamma(S^2T^*M) \end{align*} by \begin{align*} h(t)=\phi^*g(t) \end{align*} for every $t \in [0,T)$. Since each $g(t)$ is a Riemannian metric and $\phi: M \to M$ is a diffeomorphism, each $h(t)=\phi^*g(t)$ is also a Riemannian metric on $M$. The family $h(t)$ is smooth in $t$ because pullback by the fixed smooth map $\phi$ is linear and time-independent on tensor fields. At time $t=0$, using the hypothesis $\phi^*g_0=g_0$, we have \begin{align*} h(0)=\phi^*g(0)=\phi^*g_0=g_0. \end{align*} Thus $h(t)$ and $g(t)$ have the same initial metric. [/step]

[step:Verify that the pulled-back family satisfies the Ricci flow equation]Because $\phi$ is independent of $t$, time differentiation commutes with pullback: \begin{align*} \frac{\partial h}{\partial t}(t) = \frac{\partial}{\partial t}\bigl(\phi^*g(t)\bigr) = \phi^*\left(\frac{\partial g}{\partial t}(t)\right). \end{align*} Since $g(t)$ solves the Ricci flow equation, \begin{align*} \phi^*\left(\frac{\partial g}{\partial t}(t)\right) = \phi^*\bigl(-2\operatorname{Ric}(g(t))\bigr) = -2\phi^*\operatorname{Ric}(g(t)). \end{align*} The Ricci tensor is natural under pullback by diffeomorphisms: for every Riemannian metric $q$ on $M$, \begin{align*} \operatorname{Ric}(\phi^*q)=\phi^*\operatorname{Ric}(q). \end{align*} Applying this with $q=g(t)$ gives \begin{align*} \frac{\partial h}{\partial t}(t) = -2\operatorname{Ric}(\phi^*g(t)) = -2\operatorname{Ric}(h(t)). \end{align*} Therefore $h(t)$ is a Ricci flow solution on $M$ with initial metric $g_0$.[/step]

[guided]The purpose of introducing $h(t)=\phi^*g(t)$ is to turn the initial symmetry $\phi$ into a second Ricci flow candidate. Let $\Gamma(S^2T^*M)$ denote the space of smooth sections of the bundle $S^2T^*M$ of symmetric covariant $2$-tensors on $M$. Define the map \begin{align*} h: [0,T) \to \Gamma(S^2T^*M) \end{align*} by \begin{align*} h(t)=\phi^*g(t) \end{align*} for every $t \in [0,T)$. For each $t \in [0,T)$, the tensor $h(t)$ is a Riemannian metric because pullback by a diffeomorphism preserves symmetry and positive definiteness of covariant $2$-tensors. We now check the Ricci flow equation for $h(t)$. Since $\phi$ is fixed in time, pullback commutes with $\partial/\partial t$: \begin{align*} \frac{\partial h}{\partial t}(t) = \frac{\partial}{\partial t}\bigl(\phi^*g(t)\bigr) = \phi^*\left(\frac{\partial g}{\partial t}(t)\right). \end{align*} The original family $g(t)$ satisfies Ricci flow, so \begin{align*} \phi^*\left(\frac{\partial g}{\partial t}(t)\right) = \phi^*\bigl(-2\operatorname{Ric}(g(t))\bigr) = -2\phi^*\operatorname{Ric}(g(t)). \end{align*} The key geometric fact is that Ricci curvature is diffeomorphism-natural: pulling back the metric pulls back its Levi-Civita connection, its curvature tensor, and hence its Ricci contraction. Therefore \begin{align*} \operatorname{Ric}(\phi^*g(t))=\phi^*\operatorname{Ric}(g(t)). \end{align*} Substituting this identity into the previous equation gives \begin{align*} \frac{\partial h}{\partial t}(t) = -2\operatorname{Ric}(\phi^*g(t)) = -2\operatorname{Ric}(h(t)). \end{align*} Thus $h(t)$ satisfies the Ricci flow equation. Finally, the initial value agrees with that of $g(t)$ because $\phi$ is an isometry of $g_0$: \begin{align*} h(0)=\phi^*g(0)=\phi^*g_0=g_0. \end{align*} So $g(t)$ and $h(t)$ are two Ricci flow solutions on the same closed manifold, over the same time interval, with the same initial metric.[/guided]

[step:Apply uniqueness to identify the two flows] The manifold $M$ is closed. The maps \begin{align*} g: [0,T) \to \Gamma(S^2T^*M) \end{align*} and \begin{align*} h: [0,T) \to \Gamma(S^2T^*M) \end{align*} are smooth Ricci flow solutions on $M$ with the same initial metric $g_0$. By Hamilton uniqueness for Ricci flow on closed manifolds, applied on the common interval $[0,T)$, the two solutions agree: \begin{align*} h(t)=g(t) \end{align*} for every $t \in [0,T)$. The theorem lookup for Hamilton uniqueness could not be resolved in this revision environment, so the citation should be linked to the corresponding theorem entry when the database search is available. [/step]

[step:Translate equality of pulled-back metrics into preservation of isometries] Since $h(t)=\phi^*g(t)$ by definition and $h(t)=g(t)$ by uniqueness, we obtain \begin{align*} \phi^*g(t)=g(t) \end{align*} for every $t \in [0,T)$. This is precisely the statement that $\phi \in \operatorname{Isom}(M,g(t))$ for every $t \in [0,T)$. Since $\phi \in \operatorname{Isom}(M,g_0)$ was arbitrary, it follows that \begin{align*} \operatorname{Isom}(M,g_0) \subseteq \operatorname{Isom}(M,g(t)) \end{align*} for every $t \in [0,T)$. [/step]