TITLE: Exists homeomorphism which carries each fiber isomorphically to itself, composition?
QUESTION [9 upvotes]: Let $\mu$ and $\mu'$ be two different Euclidean metrics on the same vector bundle $\xi$. How do I see that there exists a homeomorphism $f: E(\xi) \to E(\xi)$ which carries each fiber isomorphically onto itself, so that the composition $\mu \circ f: E(\xi) \to \mathbb{R}$ is equal to $\mu'$?
Thoughts so far. Every positive definite matrix $A$ can be expressed uniquely as the square of a positive definite matrix $\sqrt{A}$. The power series expansion$$\sqrt{tI + X} = \sqrt{t}\left( I + {1\over{2t}} - {1\over{8t^2}}X^2 + \dots\right),$$is valid provided that the characteristic roots of $tI + X = A$ lie between $0$ and $2t$. This shows that the function $A \mapsto \sqrt{A}$ is smooth. But I am not sure how to complete. Can anyone help?
Notation. Let $B$ denote a fixed topological space, which will be called the base space. A real vector bundle $\xi$ over $B$ consists of the following:

a topological space $E = E(\xi)$ called the total space,
a (continuous) map $\pi: E \to B$ called the projection map, and
for each $b \in B$ the structure of a vector space over the real numbers in the set $\pi^{-1}(b)$.

A Euclidean vector bundle is a real vector bundle $\xi$ together with a continuous function$$\mu: E(\xi) \to \mathbb{R}$$such that the restriction of $\mu$ to each fiber of $\xi$ is positive definite and quadratic. The function $\mu$ itself will be called a Euclidean metric on the vector bundle $\xi$.

REPLY [4 votes]: There are several questions asking about the same problem on stackexchange:
Different Euclidean metrics on a vector bundle,
Isometry of two Euclidean structures on the same vector bundle,
but I haven't seen a complete correct answer. This answer https://math.stackexchange.com/a/1209444/251687 is correct but lacks some details; the others I saw were wrong.
The question is Problem 2-E in Milnor's book Characteristic Classes, and I think the author's hint is crucial, which I list as the following lemma.
Lemma. For any positive definite matrix $A$, there exists a unique positive definite matrix $B$ such that $B^2=A$. If we write $\sqrt{A}:=B$, then the map $A\mapsto\sqrt{A}$ is smooth.
Equivalently, for a positive definite matrix $A$, $\sqrt{A}$ is the unique positive definite matrix such that $\sqrt{A}^t\sqrt{A}=A$.
Now we can give two proofs of the problem. The first one is in terms of local trivializations; the second one is formulated in an intrinsic manner. 
Proof. (Define isometries locally and check they agree on overlaps.) 
By choosing local $\mu$-orthonormal basis vectors, we can cover the base manifold $M$ by $\{U_\alpha\}$ such that there are local isometric trivializations $(E|_{U_\alpha},\mu)\xrightarrow{\phi_\alpha}(U_\alpha\times\mathbb{R}^n,\mu_{\mathbb{R}^n})$, where $\mu_{\mathbb{R}^n}$ is standard metric. Let $g_{\alpha\beta}:U_\alpha\cap U_\beta\to O(n)$ be the transition maps. In each local trivialization, the inner product $((\phi_\alpha^{-1})^*\mu^\prime)|_{U_\alpha\times\mathbb{R}^n}=:\mu^\prime_{\mathbb{R}^n_\alpha}$ is given by a positive definite matrix-valued function $A_\alpha$, and $A_\alpha=g_{\alpha\beta}A_\beta g_{\alpha\beta}^{-1}$. Now define $\psi_\alpha:U_\alpha\times\mathbb{R}^n\to U_\alpha\times\mathbb{R}^n$ by $\psi_\alpha(x,v)=(x,\sqrt{A_\alpha(x)}v)$, then for any two vectors $v,w\in \mathbb{R}^n=\phi_\alpha(E_x)$, $(\psi_\alpha^*\mu_{\mathbb{R}^n})(v,w)=v^t\sqrt{A_\alpha}^t\sqrt{A_\alpha}w=v^tA_\alpha w=\mu^\prime_{\mathbb{R}^n_\alpha}(v,w)$, and hence $\mu^\prime|_{E|_{U_\alpha}}=(\phi_\alpha^{-1}\psi_\alpha\phi_\alpha)^*(\mu|_{E|_{U_\alpha}})$. So $\varphi_\alpha:=\phi_\alpha^{-1}\psi_\alpha\phi_\alpha$ gives an isometry $(E|_{U_\alpha},\mu^\prime)\to(E|_{U_\alpha},\mu).$ Next we want to check such $\varphi_\alpha=\varphi_\beta$ on overlap. (Here the notation is different from what I meant in my comment on Mike's answer.) This means $\phi_\alpha^{-1}\psi_\alpha\phi_\alpha=\phi_\beta^{-1}\psi_\beta\phi_\beta$, i.e. $\psi_\alpha=g_{\alpha\beta}\psi_\beta g_{\alpha\beta}^{-1}$. By definition of $\psi_\alpha$, what we want is $\sqrt{A_\alpha}=g_{\alpha\beta}\sqrt{A_\beta} g_{\alpha\beta}^{-1}$. But $\sqrt{A_\alpha}$ and $g_{\alpha\beta}\sqrt{A_\beta} g_{\alpha\beta}^{-1}$ are both positive definite matrices whose square equals $A_\alpha$, so by uniqueness, we are done. 
In the same spirit, we can give a more intrinsic proof.
Alternative Proof. (More intrinsic.)
We can rephrase the lemma as: Given any finite dimensional real vector space $V$ and two inner products $\mu,\mu^\prime$ on $V$, there exists a unique $\varphi\in GL(V)$, depending smoothly on $\mu,\mu^\prime$, such that

$\varphi$ is self-adjoint w.r.t. $\mu$;
$\varphi:(V,\mu^\prime)\to(V,\mu)$ is an isometry.

So now for our vector bundle $E$, one can define a unique isometry $\varphi_x:(E_x,\mu^\prime_x)\to(E_x,\mu_x)$ on each fiber $E_x$ s.t. $\varphi_x$ is self-adjoint w.r.t. $\mu_x$. This gives a global smooth isometry $\varphi:(E,\mu^\prime)\to(E,\mu)$.
Remark. The same conclusion holds true in the setting of Hermitian metrics on complex vector bundles. In other words, given a vector bundle $E\to M$, the general linear gauge group $GL(E)$ acts transitively on the space of Hermitian/Euclidean metrics on $E$. So one can study the orbit space of such action, which is $GL(E)/U(E,h_0)$ where $h_0$ is a prescribed metric. This is one way of thinking about Hermitian-Yang-Mills connections on stable holomorphic bundles.