$f:M \rightarrow N$ an isometry, show that $X$ is a killing field on $M$ if and only if $Y=df(X)$ is a killing field. The problem is stated as follows. Let $X$ be a differential vector field on $M$ and let $f: M \rightarrow N$ be an isometry. Show that the pushforward $Y=df(X)$ is a Killing field on $N$ if and only if $X$ is a Killing field on $M$.
The "only if" part is trivial once you've proved the "if" direction because $X$ is the pushforward of $Y=df(X)$ by $f^{-1}$.
I am stuck here. I am trying to prove this using the definition that a Killing field is a vector field whose flows are isometries. I'm using the notation that if $\varphi: (-\epsilon, \epsilon) \times M \rightarrow M$ is the flow of $X$ and $\psi: (-\epsilon, \epsilon) \times N \rightarrow N$ is the flow of $Y=df(X)$ (I know I'm making some minor simplifying assumptions here) then, for fixed $t_{0} \in (-\epsilon,\epsilon)$ and $p_{0}\in M$ the map
$\varphi_{t_0}(p): M \rightarrow M$ is a a diffemorphism and
$\varphi_{p_0}(t): (-\epsilon, \epsilon) \rightarrow M$ is the integral curve of $X$ starting from $p_{0}$.
Analogous statements hold for $\psi$.
Also assuming $X$ is a killing field $$\varphi_{t_0}(p): M \rightarrow M$$ is an isometry for each $t_{0}$.
I am trying to use the fact that if $\gamma: (-\epsilon, \epsilon)\rightarrow M$ is the integral curve of $X$ starting at $\p$, then  $f \circ \gamma$ is the integral curve of $Y=df(X)$ starting from $f(p)$ and the fact that the flows of $X$ are isometries but I can't seem to write it down properly.
For example, for fixed $q_0 \in N$, and writing $\hat{q}_0=f^{-1}(q)$ we have
$\psi_{q_0}(t)=f(\varphi_{\hat{q}_0}(t))$
but I can't translate this into getting that $\psi_{t_{0}}(q):N \rightarrow N$ is an isometry for each $t_{0}$.
 A: Here's a proof using the definition of Killing fields in terms of flows being isometries.  I'll use notation as you did in your post.  For a point $m\in M$, I will use the notation $\phi_m$ for the curve $\phi(\cdot, m):(-\epsilon,\epsilon)\rightarrow M$ given by mapping $(t,m)$ to $(t, f(m))$.
Lemma:  For any $m\in M$, $\psi_{f(m)} = f\circ \phi_m$.
Proof:  Let $m\in M$.  We will show that both $\psi_{f(m)}$ and $f\circ \phi_m$ are the integral curves of the vector field $Y$.  By uniqueness of these integral curves, we must then have $\psi_{f(m)} = f\circ \phi_m$.
At time $t = 0$, we have $\psi_{f(m)}(0) = f(m)$ and $f(\phi_m(0)) = f(m)$, so they agree when $t = 0$.
Now, we also have $\frac{d}{dt}|_{t=0} \psi_{f(m)} = Y_{f(m)}$ because $\psi$ is the flow of $df(X) = Y$.  On the other hand by the chain rule, we have $\frac{d}{dt}|_{t=0}(f\circ \phi_m) = df\left( \frac{d}{dt}|_{t=0} \phi_m\right) = df(X_m) = Y_{f(m)}$. $\square$
(By the way, the above lemma is does not use the Riemannian structures anywhere - it works for general manifolds!)
With this lemma established, we note that it really means that $\psi(t,f(m)) = f(\phi(t,m))$ for any $(t,m)\in (-\epsilon, \epsilon)\times M$.  This in turn implies that $\psi_t = f\circ \phi_t \circ f^{-1}$.
Now, letting $g_N$ denote the metric on $N$, and similarly for $g_M$, we have \begin{align*} \psi_{t}^\ast g_N &= (f\circ \phi_t\circ f^{-1})^\ast g_N\\ &= (f^{-1})^\ast \phi_t^\ast f^\ast g_N\\ &= (f^{-1})^\ast \phi_t^\ast g_M\\ &= (f^{-1})^\ast g_M \\ &= g_N.\end{align*}
