The answer is "NO". To show this let us use the following:

**Lemma.** Let $L$ be a lattice in $\mathbb R^q$ ($q$ is any positive integer).
Assume $$\operatorname{diam} \mathbb R^q/L>1000.$$
Then there is a midpoint $m$ of two points in $L$ such that $|m-x|>1$ for any $x\in L$?

Modulo Lemma one can construct an action of parallel translations the following way:
Let us construct inductively a sequence of lattices $L_q$ on $\mathbb R^q$ such that $\mathop{diam} \mathbb R^q/L_q<1000$ and such that $|x|>1$ for any $x\in L$.
Start with standard $L_1=\mathbb Z$ in $\mathbb R$.
To construct $L_{q}$ take
$$L_{q}'=L_{q-1}\times \mathbb Z\subset \mathbb R^{q-1}\times\mathbb R = \mathbb R^{q}.$$
If $\mathop{diam} \mathbb R^q/L'_q < 1000$ set $L_q = L'_q$.
Othewise pass to the minimal lattice which contains $L'_q$ and the midpoint provided by the Lemma.
Applying this construction finitely many times you get a lattice $L_q$ with $\mathop{diam} \mathbb R^q/L_q<1000$.

Continue the process, we get lattice $L_\infty$ in $H$ which is a $1000$-net, its fundamental doamin contains a ball of radius 1; i.e. $H/L_\infty$ is not compact.

*Proof of Lemma.*
For $z\in\mathbb R^q$, denote by $\rho(z)$ the minimal distance to a point in $L$.
Take a point $z\in\mathbb R^q$ which maximize distance to $L$.
So $\rho(z)\ge 1000$. Then there is a couple of points $x,y\in L$ such that
$\angle xzy\ge\pi/2$ and $|x-z|=|x-z|=\rho(z)$.
Let $m$ be the midpoint for $x$ and $y$.
Then
$$|z-m|\le \frac{\rho(z)}{\sqrt{2}}$$ and therefore the distance from $m$ to any point of $L$ is at least $1000{\cdot}(1-\tfrac1{\sqrt{2}})>1$. $\square$