[proofplan] We verify directly that the parabolically rescaled family satisfies the Ricci flow equation. The time derivative contributes a factor $1/\lambda$, which cancels the constant metric factor $\lambda$. The remaining point is that the Ricci tensor, viewed as a $(0,2)$-tensor, is unchanged under multiplying the metric by a positive constant. [/proofplan]

[step:Check that the rescaled family is a smooth family of Riemannian metrics] Since $\lambda > 0$, define the rescaling map \begin{align*} \rho: \lambda I &\to I, \end{align*} by the rule \begin{align*} \rho(s)=\frac{s}{\lambda}. \end{align*} This map is a smooth bijection onto $I$. Let $\operatorname{Sym}^2 T^*M \to M$ denote the vector bundle of symmetric covariant $2$-tensors on $M$, and let $\Gamma(\operatorname{Sym}^2 T^*M)$ denote its space of smooth sections. Because $g: I \to \Gamma(\operatorname{Sym}^2 T^*M)$ is smooth, the composition $g \circ \rho$ is smooth. Multiplication by the fixed scalar $\lambda$ preserves smoothness and positive-definiteness of each metric tensor, so \begin{align*} \tilde g(s)=\lambda g(s/\lambda) \end{align*} is a smooth one-parameter family of Riemannian metrics on $M$ for $s \in \lambda I$. [/step]

[step:Show that constant metric scaling leaves the Ricci tensor unchanged]Fix $t_0 \in I$, and define the Riemannian metric $h$ on $M$ by \begin{align*} h = \lambda g(t_0). \end{align*} We prove that \begin{align*} \operatorname{Ric}(h)=\operatorname{Ric}(g(t_0)). \end{align*} Let $\mathfrak{X}(M)=\Gamma(TM)$ denote the space of smooth vector fields on $M$. Let $X,Y,Z \in \mathfrak{X}(M)$ be arbitrary. Write $\nabla^g$ for the Levi-Civita connection of $g(t_0)$ and $\nabla^h$ for the Levi-Civita connection of $h$. The Koszul formula gives \begin{align*} 2h(\nabla^h_XY,Z)=X(h(Y,Z))+Y(h(Z,X))-Z(h(X,Y))-h(X,[Y,Z])+h(Y,[Z,X])+h(Z,[X,Y]). \end{align*} Since $h=\lambda g(t_0)$ and $\lambda$ is constant on $M$, the right-hand side is \begin{align*} \lambda\bigl(2g(t_0)(\nabla^g_XY,Z)\bigr). \end{align*} The left-hand side is \begin{align*} 2h(\nabla^h_XY,Z)=2\lambda g(t_0)(\nabla^h_XY,Z). \end{align*} Because $\lambda>0$, cancellation gives \begin{align*} g(t_0)(\nabla^h_XY,Z)=g(t_0)(\nabla^g_XY,Z). \end{align*} The vector field $Z$ was arbitrary and $g(t_0)$ is nondegenerate, hence \begin{align*} \nabla^h_XY=\nabla^g_XY. \end{align*} Thus the two Levi-Civita connections agree. Let $R^h$ and $R^g$ denote the curvature operators of $\nabla^h$ and $\nabla^g$, defined for vector fields $X,Y,Z \in \mathfrak{X}(M)$ by \begin{align*} R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z. \end{align*} Since the connections agree, their curvature operators agree: \begin{align*} R^h(X,Y)Z=R^g(X,Y)Z. \end{align*} The Ricci tensor is the trace of the linear endomorphism $V \mapsto R(V,X)Y$, so this equality of curvature operators gives \begin{align*} \operatorname{Ric}(h)(X,Y)=\operatorname{Ric}(g(t_0))(X,Y). \end{align*} Because $X$ and $Y$ were arbitrary smooth vector fields, the tensor identity \begin{align*} \operatorname{Ric}(h)=\operatorname{Ric}(g(t_0)) \end{align*} holds on all of $M$.[/step]

[guided]The only geometric input needed is the behaviour of Ricci curvature under multiplying the metric by a positive constant. We prove it through the Levi-Civita connection rather than through indexed coordinate components. Fix $t_0 \in I$, and define a new metric $h$ on $M$ by \begin{align*} h=\lambda g(t_0). \end{align*} Let $\mathfrak{X}(M)=\Gamma(TM)$ be the space of smooth vector fields on $M$, and choose arbitrary vector fields $X,Y,Z \in \mathfrak{X}(M)$. Write $\nabla^g$ for the Levi-Civita connection of $g(t_0)$ and $\nabla^h$ for the Levi-Civita connection of $h$. The Koszul formula characterizes the Levi-Civita connection from the metric and Lie brackets of vector fields. Applied to $h$, it says \begin{align*} 2h(\nabla^h_XY,Z)=X(h(Y,Z))+Y(h(Z,X))-Z(h(X,Y))-h(X,[Y,Z])+h(Y,[Z,X])+h(Z,[X,Y]). \end{align*} Now use the fact that $h=\lambda g(t_0)$ with $\lambda$ constant in space. Every term on the right-hand side acquires exactly one factor of $\lambda$; for instance, \begin{align*} X(h(Y,Z))=X(\lambda g(t_0)(Y,Z))=\lambda X(g(t_0)(Y,Z)). \end{align*} The same constant factor appears in the remaining five terms. Therefore the right-hand side of the Koszul formula for $h$ is \begin{align*} \lambda\bigl(2g(t_0)(\nabla^g_XY,Z)\bigr), \end{align*} where the expression in parentheses is the Koszul formula for the metric $g(t_0)$. The left-hand side also contains the metric $h$, hence \begin{align*} 2h(\nabla^h_XY,Z)=2\lambda g(t_0)(\nabla^h_XY,Z). \end{align*} Combining the two identities and cancelling $2\lambda$, which is allowed because $\lambda>0$, gives \begin{align*} g(t_0)(\nabla^h_XY,Z)=g(t_0)(\nabla^g_XY,Z). \end{align*} Since $Z$ was arbitrary and $g(t_0)$ is nondegenerate, the two vector fields paired against $Z$ must be equal: \begin{align*} \nabla^h_XY=\nabla^g_XY. \end{align*} This proves that multiplying a metric by a positive constant does not change its Levi-Civita connection. Next define $R^h$ and $R^g$ to be the curvature operators of $\nabla^h$ and $\nabla^g$. For a connection $\nabla$, its curvature operator is the map determined by \begin{align*} R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z. \end{align*} Because this expression uses only the connection and the Lie bracket, the equality $\nabla^h=\nabla^g$ implies \begin{align*} R^h(X,Y)Z=R^g(X,Y)Z \end{align*} for all smooth vector fields $X,Y,Z$. The Ricci tensor is obtained by taking the trace of the linear endomorphism $V \mapsto R(V,X)Y$. Since the underlying curvature operators are equal, their traces are equal, and hence \begin{align*} \operatorname{Ric}(h)(X,Y)=\operatorname{Ric}(g(t_0))(X,Y). \end{align*} The vector fields $X$ and $Y$ were arbitrary, so this proves the global tensor equality \begin{align*} \operatorname{Ric}(h)=\operatorname{Ric}(g(t_0)). \end{align*}[/guided]

[step:Differentiate the rescaled metric and obtain the Ricci flow equation] Let $s \in \lambda I$, and define $t=s/\lambda \in I$. If $s$ is an interior point of $\lambda I$, the following derivative is the ordinary time derivative; if $s$ is an endpoint of $\lambda I$, it is the corresponding one-sided derivative, and $t$ is the corresponding endpoint of $I$. Applying the ordinary or one-sided chain rule, as appropriate, to \begin{align*} \tilde g(s)=\lambda g(s/\lambda), \end{align*} we get \begin{align*} \partial_s \tilde g(s) = \lambda \cdot \frac{1}{\lambda}\partial_t g(t) = \partial_t g(t). \end{align*} Since $g(t)$ satisfies the Ricci flow equation, \begin{align*} \partial_t g(t)=-2\operatorname{Ric}(g(t)). \end{align*} By the constant-scaling invariance proved above, applied to the metric $g(t)$ and the scalar $\lambda$, \begin{align*} \operatorname{Ric}(\tilde g(s)) = \operatorname{Ric}(\lambda g(t)) = \operatorname{Ric}(g(t)). \end{align*} Combining these identities gives \begin{align*} \partial_s \tilde g(s) = -2\operatorname{Ric}(g(t)) = -2\operatorname{Ric}(\tilde g(s)). \end{align*} This holds for every $s \in \lambda I$, so $(M,\tilde g(s))_{s \in \lambda I}$ is a Ricci flow. [/step]