Setup: Let $f:\mathbb R^3\to\mathbb R^3$ be a similarity transformation. Then $f=rA+b$ for some fixed orthogonal matrix $A$, vector $b$ and nonzero real $r$. Suppose $S$ is a surface, and $S'=f(S)$. If $N$ is a Gauss map on $S$, and $N'\circ f=A\circ N$, then $N'$ is a Gauss map on $S'$. Let $K,H,K',H'$ denote the Gauss and mean curvatures of $S$ and$ S'$ respectively.
Claim: $K'\circ f=\frac K{r^2}\quad$ and $\quad H'\circ f=\frac Hr$
Idea: For any $v,w\in T_pS$, we have $\langle df_p(v),df_p(w)\rangle=r^2\langle v,w\rangle$. Moreover, if $\sigma'$ and $\sigma$ are the second fundamental forms of $S'$ and $S$ respectively, then $r\sigma'_{f(p)}(Av,Aw)=\sigma_p(v,w)$
The claim should immediately follow from these equalities.
Problem: Prove the equalities and conclude.
Attempt: $\langle df_p(v),df_p(w)\rangle=\langle rAv,rAw\rangle=r^2\langle Av,Aw\rangle=r^2\langle v,w\rangle$, where the last equality follows from the orthogonality of $A$.
$$r\sigma'_{f(p)}(Av,Aw)=-r\langle dN'_{f(p)}(Av),Aw\rangle=-r\langle AdN_p(df^{-1}_{f(p)}(Av)),Aw\rangle=$$ $$-\langle AdN_p(v),Aw\rangle=-r^{-2}\langle df(dN_p(v)),df(w)\rangle=-\langle dN_p(v),w\rangle=\sigma_p(v,w)$$
Now, $$K'\circ f(p)=\det dN'(f(p))=\det [AdN(p)df^{-1}_{f(p)}]=\frac 1r \det [AdN(p)A^{-1}]=$$ $$\frac 1r \frac{\det A}{\det A}\det [dN(p)]=\frac 1r \det [dN(p)]=\frac 1r K(p)$$
This equality is not as expected.
$H'\circ f(p)=-\frac 12 tr(dN'(f(p)))=-\frac 1{2r} tr(AdN_p A)=-\frac 1{2r} tr(A^2dN_p)$ How can we continue?
Attempt 2: Take $\{e_1,e_2\}\subset T_pS$ to be principal directions of $S$, and $k_i$ to be the principle curvatures. Then $dN'_{f(p)}(Ae_i)=AdN_p[\frac 1r A^{-1}(Ae_i)]=-\frac 1r k_i Ae_i$. Hence $\{Ae_1,Ae_2\}\subset T_{f(p)}S'$ are principal directions of $S'$, and the principle curvatures are $\frac 1r k_i$. The conclusion follows.
I am supposed to use the Idea but I do not see how. Let us remedy the first attempt.
In the third equality of the Gaussian curvature calculation, you appear to have $\det(\frac{1}{r}A) = \frac{1}{r} \det(A)$. But $A$ is $2 \times 2$, and $\det(cA) = c^{n} \det(A)$ if $A$ is $n \times n$. :)
Separately, $df_{f(p)}^{-1} = \frac{1}{r} A^{-1}$ (rather than $\frac{1}{r}A$). This doesn't affect the outcome of the Gaussian curvature computation, since $\det(A^{-1}) = 1$, but it does have some bearing on the mean curvature.