Poisson's equation under translation and scaling Let $u\in C^\infty (\Omega,\mathbb R)$ be a solution of 
$$\begin{array}{cccl} -\Delta u & = & 0 & \mathrm{in}\ \Omega \\
u & = & g & \mathrm{on}\ \partial\Omega\end{array}$$
Whereas $\Omega\subset\mathbb R^n$.
We define $\delta_\lambda(x):=\lambda x$ and $\tau_\lambda (x):= x+c$. Find PDEs and domains $\tilde\Omega_1, \tilde\Omega_2$ with solutions $u_\lambda :=u\circ \delta_\lambda$ and $u_c:=u\circ\tau_c$ respectively.
I think by introducing $x'=x+c$ we have
$$\frac{\partial u(x')}{\partial x} = \frac{\partial u}{\partial x'}\frac{\partial x'}{\partial x} = \frac{\partial u}{\partial x'},$$
so $u\circ\tau_c$ is also a solution of $-\Delta u =0$? I'd say we just have to change our domain to $\tilde\Omega_1:=\Omega+c:=\{x+c|x\in\Omega\}$. 
Also we set $y'=\lambda x$ and get
$$\frac{\partial u(y')}{\partial x} = \frac{\partial u}{\partial y'}\frac{\partial y'}{\partial x} = \lambda \frac{\partial u}{\partial y'}$$
so $u\circ\delta_\lambda$ is a solution of $-\frac{1}{\lambda^2}\Delta u=0$ on $\tilde\Omega_2:=\lambda\Omega:=\{\lambda x|x\in\Omega\}$.
Am I going in the right direction with this? 
 A: The domains like I said in the comments above
$$
u\circ \tau_c :\widetilde{\Omega}_1 \to \mathbb{R},
$$
knowing $u: \Omega\to \mathbb{R}$, we must have
$$
\tau_c : \widetilde{\Omega}_1 \to \Omega,
$$
this is to say,
$$
\Omega = \tau_c(\widetilde{\Omega}_1),
$$
applying the inverse of the translation operator $\tau_{-c}$ on both sides:
$$
\widetilde{\Omega}_1 = \tau_{-c}{\Omega} = \{x'\in \mathbb{R}^n : x' = x-c\text{ for some }x\in \Omega\}.
$$
Similarly $\widetilde{\Omega}_2 = \delta_{1/\lambda}{\Omega}$. Both are the same with what you wrote in the comments. Following this kind of notation will get you through more complicated coordinate change.
Also here is a comment for the differentiation you have in your question: notice in the new domain $\widetilde{\Omega}_1$, the differentiation should be with respect to $x'\in \widetilde{\Omega}_1$, not with respect to the original $x$ (just to avoid confusion of notations), the partial derivative should be:
$$
\frac{\partial (u\circ \tau_c)(x')}{\partial x'} = \frac{\partial u(x)}{\partial x}\Bigg|_{x = \tau_c(x')}\cdot \frac{\partial x}{\partial x'}\Bigg|_{x = \tau_c(x')} = \frac{\partial u(x)}{\partial x}\Bigg|_{x = \tau_c(x')},
$$
the new equation should be letting $(u\circ \tau_c)(x') = u_{c}$
$$
-\Delta_{x'} u_{c} = 0, \quad \text{ in } \widetilde{\Omega}_1,\\
u_{c}(x') = (g\circ \tau_c)(x') \quad \text{ on } \partial \widetilde{\Omega}_1.
$$
Now do the same thing for the translation operator.
