Let me write $G$ instead of $X$ since we're talking about compact groups and my groups refuse to be called $X$...
Silliness aside, the point here is that for each $x \in G$ the map $\tau_x\colon L^q(G) \to L^q(G)$ is an invertible (in fact isometric) linear map. You get a group homomorphism $\tau\colon G \to {\rm GL}(L^q(G))$ and as such $\tau$ is strongly continuous (provided $1 \leq q \lt \infty$). What this means is that for each $f \in L^q(G)$ the function $x \mapsto \tau_x f$ is continuous as a function $(G,d) \to (L^q(G),\lVert\cdot\rVert_q)$, which is precisely the continuity Thomas mentions in his answer.
The reason this is the case is that the space of continuous functions $C(G)$ is dense in $L^q(G)$ and that continuous functions on compact spaces are uniformly continuous. For (uniformly) continuous functions $f\colon G \to \mathbb{C}$ we have that $x \mapsto \lVert \tau_x f - f\rVert_\infty$ is continuous and since $\mu(G) = 1$ we have that $\lVert f \rVert_q \leq \lVert f\rVert_\infty$, so for continuous functions $f$ the map $x \mapsto \tau_x f$ is continuous with respect to all $L^q$-norms.
Now, for continuous functions it is clear that $\lVert \tau_x f\rVert_q = \lVert f\rVert_q$, so $\tau_x\colon (C(G),\lVert\cdot\rVert_q) \to (C(G),\lVert \cdot \rVert_q)$ is an isometry and thus it extends uniquely to an isometry of the completion $L^q(G)$ of $(C(G),\lVert \cdot \rVert_q)$. Since $\tau_{-x} \tau_x = {\rm id}_{C(G)} = \tau_{x} \tau_{-x}$ we see that each of those extensions is invertible. To see that the resulting map $\tau\colon G \to {\rm GL}(L^q(G))$ is strongly continuous, let $h \in L^q(G)$ and $\varepsilon \gt 0$. Then there is $f \in C(G)$ such that $\lVert f - h\rVert_q \lt \varepsilon$ and by continuity of $x \mapsto \lVert\tau_x f - f\rVert_\infty$ there is $\delta$ such that $d(x,e) \lt \delta$ implies that $\lVert\tau_x f - f\rVert_\infty \lt \varepsilon$. But then $ \begin{align*} \lVert \tau_x h - h\rVert_q & \leq \lVert \tau_x h - \tau_x f\rVert_q + \lVert \tau_x f - f\rVert_q + \lVert f - h\rVert_q \\ &\leq \lVert \tau_x h - \tau_x f\rVert_q + \lVert \tau_x f - f\rVert_\infty + \lVert f - h\rVert_q \\ & \lt 3 \varepsilon \end{align*} $ where we have used that $\lVert \tau_x h - \tau_x f\rVert_q = \lVert h - f\rVert_q$ by translation invariance of Haar measure.
To see that $q \lt \infty$ is essential, consider the characteristic function of a non-trivial segment $[0,\alpha]$ of the circle group $S^1$.
Added:
All I said here extends (with small modifications) to non-commutative locally compact groups, see the threads