Isometry group of a norm is always contained in some Isometry group of an inner product? $\newcommand{\<}{\langle} \newcommand{\>}{\rangle} $Let $||\cdot||$ be a norm on a finite dimensional real vector space $V$.
Does there always exist some inner product $\<,\>$ on $V$ such that $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$ ? 
Update:
As pointed by Qiaochu Yuan the answer is positive.
This raises the question of uniqueness of the inner product $\<,\>$ which satisfies $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$. 
Is it unique (up to scalar multiple)?
Remarks:
1) Determining $\<,\>$ (up to scalar multiple) is equivalent to determining $\text{ISO}(\<,\>)$. 
Clearly if we know the inner product we know all its isometries. The other direction follows as a corollary from an argument given here which shows which inner products are preserved by a given automorphism. 
2) Since there are "rigid" norms (whose only isometries are $\pm Id$ ) the uniqueness certainly doesn't hold in general.
One could hope for that in the case of "rich enough norms" (norms with many isometries, see this question) the subset $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$ will be large enough to determine $\text{ISO}(\<,\>)$. 
(which by remark 1) determines $(\<,\>)$).  
 A: $\newcommand{\<}{\langle} \newcommand{\>}{\rangle} $
For completeness, I am writing more detailes of the solution suggested by
Qiaochu:
Denote by $G$ the isometry group of $(V,\|\cdot\|)$. $G\subseteq \text{End}(V)$. On $\text{End}(V)$  we have the operator norm $\| \|_{op}$ (w.r.t the given norm $\|\cdot\|$), which induces a topology on $\text{End}(V)$. 
Lemma 1: $G$ is compact in $\text{End}(V)$
Proof: $\text{End}(V)$ is a finite dimensional normed space, hence it is linearly homeomorphic to $\mathbb{R}^n$ (This is in fact true for every finite dimensioanl real topological vector space). So, the Heine-Borel theorem aplies. (Every closed and bounded subset is compact).
$G$ is bounded since for every isometry $g\in G, \|g\|_{op}=1$, hence $G$ is contained in the unit sphere of $(\text{End}(V),\| \|_{op})$ .
$G$ is closed: Assume $g_n \rightarrow   g,g_n\in G$. Fix some $v\in V$. $\|g_n(v)-g(v)\|_V\leq \|g_n-g\|_{op}\cdot\|v\|_V \xrightarrow{n\to\infty} 0$. 
So $g_n(v)\xrightarrow{n\to\infty}g(v)$. Now by the continuity of the norm $\| \cdot \|$ (w.r.t to the topology it induces on $V$) we get that:
$\|v\| \stackrel{g_n isometry}{=} \|g_n(v)\|\xrightarrow{n\to\infty}\|g(v)\|$. This forces $g$ to be an isometry.
Corollary1: $G$ is locally compact Hausdorff topological group.
Proof: $\text{End}(V)$ is Hausdorff (Any metric space is...), and every subspace of Hausdorff is also Hausdorff. It is sdandard fact that $GL(V)$ is a topological group (t.g), and any subgroup of a t.g is a t.g.

Now, there exists a left-invariant measure $\mu$ on the Borel $\sigma$-algebra of $G$ such that $\mu(G)>0$. (This is the Haar measure which can be constructed on any locally compact Hausdorff topological group).
Now take any inner product $\<,\>$ on $V$. Fix $v,w \in V$. Define $f_{v,w}:G\rightarrow \mathbb{R},f_{v,w}(g)=\<gv,gw\>$.
Lemma 2: $f_{v,w}$ is continuous
Proof: Since $G$ is a metric space (a subspace of the normed space $\text{End}(V)$) it is enough to check sequential continuity. Take
$g_n \rightarrow   g,g_n\in G$. We already showed this implies $g_n(v)\xrightarrow{n\to\infty}g(v)$ so by the continuity of the inner product $f_{v,w}(g_n)=\<g_nv,g_nw\> \xrightarrow{n\to\infty} f_{v,w}(g) $.
In particular $f_{v,w}$ is measurable, so we can integrate it. (Compactness of $G$ implies $f_{v,w}$ is bounded, and $G$ being a finite measure space guarantees the integral will be finite).
So we define: $\<v, w \>' = \int_G  f_{v,w} \, d\mu = \int_G \< gv, gw \> \, d\mu$.
Now all is left is to show $\<,  \>'$ is an inner product on $V$ that is presrved by each $h\in G$. (since this means $G=\text{ISO}(V,\|\cdot\|) \subseteq \text{ISO}(V,\<,  \>')$ as required.
Lemma 3: $\<,\>'$ is an inner product.
The only non-trivial thing is positive-definiteness. (The rest follows from the linearity of $g\in G$ and the integral, and the bilinearity of $\<,\>$). But this follows from standard measure theory:
Fix $v\neq 0$. $f_{v,v} > 0$ on $G$ (since each $g\in G$ is injective and the original inner prodcut is positive).
But this forces $\<v,v\>'>0$ as required. 
Lemma 4: $\<,  \>'$ is $G$-invariant. 
But this follows from another standard proposition in measure theory:
(See "Real Analaysis" by H.L.Royden) chapter 22 pg 488  proposition 10).
A: Yes. This is because an isometry group is always compact (with respect to the topology on $\text{End}(V)$ induced by the operator norm: this is a consequence of the Heine-Borel theorem). Hence you can average an inner product over it with respect to Haar measure.
