Perhaps the most clear way to see what's going on is to compare the action $$ S_1[\phi]=\int_{\mathbb R^d} \phi(x)^2\mathrm dx\tag1 $$ to the action $$ S_2[\phi]=\int_{\mathbb R^d} x^2\phi(x)^2\mathrm dx\tag2 $$
The first one should be invariant under translations, while the second one should not.
We define the translation operator $T_a\in\mathrm{End}(\mathscr C(\mathbb R^d))$, with $a\in\mathbb R^d$, as $$ (T_a\phi)(x):=\phi(x-a)\tag3 $$
With this, an action is invariant under translations if and only if $$ S[T_a\phi]\equiv S[\phi],\quad\forall a\in\mathbb R^d\tag4 $$
What follows is a standard textbook exercise: $$ S_1[T_a\phi]\overset{(1)}=\int_{\mathbb R^d} (T_a\phi)(x)^2\mathrm dx\overset{(3)}=\int_{\mathbb R^d} \phi(x-a)^2\mathrm dx=\int_{\mathbb R^d} \phi(y)\mathrm dy=S_1[\phi]\tag5 $$ and $$ S_2[T_a\phi]\overset{(2)}=\int_{\mathbb R^d} x^2(T_a\phi)(x)^2\mathrm dx\overset{(2)}=\int_{\mathbb R^d} x^2\phi(x-a)^2\mathrm dx=\int_{\mathbb R^d} (y+a)^2\phi(y)^2\mathrm dy\neq S[\phi]\tag6 $$ where in both cases we defined $y:=x-a$. This is the expected result: $S_1$ is translation invariant, and $S_2$ is not.