Suppose you have a convex polygon $P=\mathrm{conv}(\{(x_1,y_1),\dots, (x_k,y_k)\})$ and you stretch it in one dimension, that is, we choose $\alpha>1$ and get a new polygon $P^\alpha=\mathrm{conv}(\{(\alpha x_1,y_1),\dots, (\alpha x_k,y_k)\})$.
Is it true that you can translate and rotate $P$ to make it fit into $P^\alpha$?
This seems true to me, as you are somehow making it bigger and "keeping the shape", but I have no additional insight in how one could prove such a thing.
Yes, this is always possible.
Sandwich $P$ between two lines of some variable slope $p$, touching $P$ at $A$ and $B$ respectively, where $AB$ has slope $q$. Without loss of generality, assume $q > 0$ when $p = 0$. We can vary this configuration continuously such that $p$ increases from $0$ while $q$ increases toward $\infty$, so $pq$ will increase continuously from $0$ toward $\infty$, while $p < q$ at all times. (The one exception is if $AB$ becomes vertical while $p = 0$, but then we’re already done: we can fit $P$ into $P^\alpha$ without any rotation.) Stop when $pq = \alpha$.
We claim that rotating $P$ clockwise about $A$ by $\theta = \tan^{-1} \frac{pq - 1}{p + q}$, then translating it by $\vec{AA^\alpha}$, sends it to a subset of $P^\alpha$.
To see this, it suffices to show that the linear transformation about origin $A$ composed of this rotation by $\theta$ followed by a stretching in the $x$ direction by ratio $\frac{1}{\alpha}$:
$$\begin{bmatrix} \frac1\alpha \cos \theta & \frac1\alpha \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} = \frac{1}{\sqrt{(1 + p^2)(1 + q^2)}}\begin{bmatrix} \frac{p + q}{pq} & \frac{pq - 1}{pq} \\ 1 - pq & p + q \end{bmatrix}$$
sends $P$ to a subset of itself. You can verify that this transformation has eigenvectors
$$v_1 = \begin{bmatrix}1 \\ p\end{bmatrix}, v_2 = \begin{bmatrix}1 \\ q\end{bmatrix},$$
with eigenvalues
$$\lambda_1 = \frac1p \sqrt{\frac{1 + p^2}{1 + q^2}}, \lambda_2 = \frac1q \sqrt{\frac{1 + q^2}{1 + p^2}},$$
where $\lambda_1 < \lambda_2 < 1$ because
$$\begin{gather*} \lambda_2 - \lambda_1 = \frac{(q - p)(pq - 1)}{pq\sqrt{(1 + p^2)(1 + q^2)}} > 0, \\ 1 - \lambda_2^2 = \frac{(pq)^2 - 1}{(1 + p^2)q^2} > 0. \end{gather*}$$
So we can reinterpret this transformation as the composition of a stretching along each line of slope $p$ about its intersection with $AB$ by ratio $\frac{\lambda_1}{\lambda_2}$, followed by a dilation about $A$ by ratio $\lambda_2$; both these component transformations send $P$ to a subset of itself because it’s convex. This proves the claim.
I think this hipothesis not true.
With a simple transformation we can make the stretch direction as parallel to X-axis. So, I draw two consecutive edges $A,B,C$ of a convex polygon, let's say they are at "north", and do a X-scale to get $D,E,F$:
As you can see, the original $ABC$ must be translated to $UEV$, by an horizontal distance $BE=d=\alpha·B_x-B_x=(\alpha-1)B_x$. It must be also rotated so $U,V$ be inner to $DEF$
The problem arises when you want to apply the same translation to the rest of vertices.
$BE$ must be the greatest distance of all possible-applied translations, otherwise other vertices will fall outside of the stretched polygon. This itself does not seem a great issue, just choose the biggest one. But it may be that that chosen distance is "$d_{max}>>d$; in this case translating $B$ with $d_{max}$ will make it to fall at the very right of $E=\alpha·B$
To get things worst, each translation has an associated rotation. The likelihood of any $translation, rotation$ pair fitted for one vertex, to be valid for the entire polygon is very low, zero in general.
It's also obvious that there are cases where such a pair can be obtained. For example the triangle in the image above.
