Riemannian connection under isometry

Question

I need help completing a certain proof.
A Diffeomorphism $\varphi: M \to \tilde{M}$ of smooth manifolds induces an isomorphism. $\tilde{M}$ has an affine connection $\tilde{\nabla}$. Define a vector field $\nabla_{X}Y \in \mathfrak{X}(M)$.by $\varphi_{*}(\nabla_{X}Y)=\tilde{\nabla}_{\varphi_{*}X}(\varphi_{*}Y)$. Suppose $\varphi$ is an isometry of Riemannian manifolds and $\tilde{\nabla}$ is a Riemannian connection. Show that the connection $\nabla$ is the Riemannian Connection on $M$.
I have part of a proof that $\nabla$ agrees with the metric, but I can't seem to get the punchline. Here's what I have.
$$\varphi_{*}Z \left<\varphi_{*}X,\varphi_{*}Y \right>=\left< \tilde{\nabla}_{\varphi_{*}Z}(\varphi_{*} X),\varphi_{*}Y \right>+ \left< \varphi_{*} X,\tilde{\nabla}_{\varphi_{*}Z}(\varphi_{*}Y) \right> $$ By definition of $\nabla$ this equation becomes $$\varphi_{*}Z \left<\varphi_{*}X,\varphi_{*}Y \right>=\left<\varphi_{*}(\nabla_{Z}(X)),\varphi_{*}Y \right>+ \left< \varphi_{*} X,\varphi_{*}(\nabla_{Z}(Y) )\right> $$ Since $\varphi$ is an isometry we know that $\left<\varphi_{*}(\nabla_{Z}(X)),\varphi_{*}Y \right>= \left<\nabla_{Z}(X),Y \right>$ and $\left< \varphi_{*} X,\varphi_{*}(\nabla_{Z}(Y) )\right>=\left< X,\nabla_{Z}(Y) \right>$ therefore this equation becomes $$\varphi_{*}Z \left<X,Y \right>=\left<\nabla_{Z}(X),Y \right>+ \left< X,\nabla_{Z}(Y) \right> $$. This is where I'm stuck because I have that $\varphi_{*}Z \left<X,Y \right>$ where i need $Z \left<X,Y \right>$. Any input would be appreciated.