control of the $C^{1}$ norm of a diffeomorphism Let $\mathcal{E}$ be the set of smooth manifolds with boundary $E\subset \mathbb{R}^{3}$ which are perturbations of the unit  ball whose volume $V$, diameter $d$ and area of the boundary $A$ satisfy:
$$1\leq d\leq 4,$$
$$2\pi \leq A \leq 8\pi$$
and
$$\frac{2\pi}{3}\leq V \leq \frac{8\pi}{3}.$$
Additional assumption: The total curvature, $\int_{\partial E} H^2 d\sigma$, of the boundary is smaller $8\pi$.
I am looking for a reference where I can find a theorem which states that there exists a universal constant $C$ such that for every $E\in\mathcal{E}$ that is diffeomorphic to the closed unit ball, we can find a representative diffeomorphism with $C^{1}$ norm bounded by $C$. 
 A: The curvature assumption looks unnatural when your goal is a $C^1$ bound. Ideally, $E$ should be merely $C^1$-smooth. But here is what I can prove: 

Claim If $E$ is generated by rotating a $C^{1,\alpha}$-smooth closed curve $\Gamma$ (where $\alpha>0$), then such a diffeomorphism exists, with constant $C$ that depends only on the chord-arc constant of $\Gamma$. 

A closed curve is called chord-arc (or Lavrentiev curve) if there is a constant $M$ such that for every two points
$a,b$ on the curve the shorter arc between them has length at most $M|a-b|$, where $|a-b|$ is the Euclidean distance. 
In other words, subarcs cannot be much longer then the corresponding chords. 
Sketch of Proof.  We may and do assume that the axis of rotation is the $z$-axis. 
Then $\Gamma$ is a symmetric curve in the $xz$-plane which is symmetric about $z$-axis: formally, $\rho(\Gamma)=\Gamma$ where $\rho(x,z)=(-x,z)$. 
Step 1. It suffices to find a diffeomorphism $g$ of the $xz$-plane such that $Dg$ and $D(g^{-1})$ are bounded by suitable $C$, and $g\circ \rho=\rho\circ g$. 
Indeed, such $g$ naturally yields a diffeomorphism $G$ of $\mathbb R^3$ which, in terms of cylindrical coordinates $(r,\theta,z)$, leaves $\theta $ unchanged and applies $g$ to the pair $(r,z)$. A direct computation of derivatives shows that the derivative of $G$ also admits a two-sided bound. The key point here is that $g$ maps 
the $z$-axis to itself, which in view of the derivative bounds implies that the $x$-coordinate of $g(x,z)$ is comparable
to $x$. Because of this fact, the transition from $2$ to $3$ dimensions does not introduce much extra distortion. 
Step 2. The following theorem can be found in several places with different proofs: Theorem 7.10 in Pommerenke, also [Tukia] and [Jerison-Kenig]. 

Theorem A. Let $f:S^1\to\mathbb R^2$ be a bi-Lipschitz embedding: that is, $$M^{-1}|a-b|\le |f(a)-f(b)|\le M |a-b| \ \text{ for all } \ a,b\in S^1\tag{1}$$ where $M$ is a  positive constant. Then $f$ extends to a homeomorphism
   $F:\mathbb R^2\to\mathbb R^2$ which is bi-Lipschitz with a constant $C=C(M)$. 

The proof of Theorem A in [Pommerenke] follows the original paper by Tukia, but uses the Douady-Earle extension 
instead of Beurling-Ahlfors. Let $G$ be the domain bounded by $\Gamma$. Let $\phi:\mathbb D\to G$ be a conformal map 
(note that we can choose $\phi$ appropriately symmetric when $G$ is symmetric). The map $\phi$ extends to a homeomorphism
of the boundaries. We consider $\phi^{-1}\circ f$, which is a homeomorphism of $\partial \mathbb D$ onto itself. 
The Douady-Earle extension provides a nice (quasiconformal, smooth) diffeomorphism $\psi:\mathbb D\to\mathbb D$ 
which agrees with $\phi^{-1}\circ f$ on the boundary. Then $F=\phi\circ \psi$ is our extension. (Or rather half of it;
the process then repeats on the exterior of $\Gamma$, but you do not need this.) 
By construction $F$ is smooth on $\mathbb R^2\setminus S^1$ and therefore satisfies the desired derivative bounds there (the  bi-Lipschitz property does not imply differentiability, but controls the size of derivatives when they 
do exist). Furthermore, $F$ has the desired symmetry property $F\circ \rho=\rho\circ F$. 
Step 3. It remains to discuss 
the smoothness of $F$ up to the boundary $\partial \mathbb D$. 
One has to be  careful here because in general, a conformal map onto a $C^1$ domain is not $C^1$ up to the boundary. However, if we assume a little extra smoothness - $C^{1,\alpha}$ with $\alpha>0$ - then $\phi:\mathbb D\to G$ is in
$C^{1,\alpha}(\overline{\mathbb D})$, and $|f'|$ is bounded away from zero. This is a well-known theorem (Kellogg, Warschawski) which can be found in section 3.3 of the aforementioned book by Pommerenke, or in section II.4 of 
Garnett-Marshall. Therefore, 
$\phi^{-1}\circ f$ is a $C^1$-diffeomorphism of the circle. 
Theorem 2 in [Hu-Pal] tells us that the Douady-Early extension of a $C^1$-diffeomorphism of the circle has 
continuous derivatives up to the boundary. (Interestingly, the webpage of S. Pal indicates that they are working on 
$C^{1,\alpha}$ case now.) This completes the puzzle: since the derivative  of $\psi$ is continuous on the 
closed disk, the map $F=\phi\circ \psi$ has the same property. $\Box$ 
References (in addition to embedded links to the books): 


*

*[Hu-Pal] Jun Hu and Susovan Pal, Boundary differentiability of Douady-Earle extensions of diffeomorphisms of $S^n$, preprint available.

*[Jerison-Kenig] Jerison, David S.; Kenig, Carlos E. Hardy spaces, $A_{\infty }$, and singular integrals on chord-arc domains. Math. Scand. 50 (1982), no. 2, 221--247. MR0672926 (84k:30037) 

*[Tukia] Tukia, Pekka. Extension of quasisymmetric and Lipschitz embeddings of the real line into the plane. Ann. Acad. Sci. Fenn. Ser. A I Math. 6 (1981), no. 1, 89--94. MR0639966 (83d:30022)


Still, it would be nice to have a reference (or proof) for the following quantitative $C^1$ Schoenflies theorem, 
without the $C^{1,\alpha}$ assumption. 

Theorem/Conjecture B. Let $f:S^1\to\mathbb R^2$ be a $C^1$ embedding. Let $M$ be a constant such that $|f'|\le M$ and $|f(a)-f(b)|\ge M^{-1} |a-b|$ for all $a,b\in S^1$. Then $f$ extends 
  to a diffeomorphism $F:\mathbb R^2\to\mathbb R^2$ such that $DF$ and $D(F^{-1})$ are bounded by a constant $C=C(M)$. 


(Counterexample to an earlier version of the question): let $E$ be the union of the unit ball $B=\{x^2+y^2+z^2\le 1\}$ with the cylinder $C=\{x^2+y^2\le \epsilon^2, 0\le z\le 2\}$ (this is not a smooth manifold, but can be easily smoothed). The boundary of $E$ contains a cylinder-shaped region (topological disk) $R$ of area $\approx 2\pi \epsilon$ bounded by a curve $\partial R$ of length about $2\pi \epsilon$. If $f:E\to B$ is a diffeomorphism with derivative bounded by $C$, then  $f(\partial R)$ has length $\le 2\pi C\epsilon$. The curve $f(\partial R)$ separates $\partial B$ into two regions, the smaller of which has area of order $\epsilon^2$ by the isoperimetric inequality. It follows that $f^{-1}$ maps a region of area $O(\epsilon^2)$ onto a region of area $\ge 2\pi \epsilon$, which means that the derivative of $f^{-1}$ cannot be bounded by a universal constant. 
