The problem
Consider a complete separable metric space $(M,d)$. The group of isometries $\operatorname{Iso}(M)$ of $M$ can be endowed with topology of pointwise convergence, making it a Polish group. Let $G$ be a subgroup of $\operatorname{Iso}(M)$. With the induced topology $τ$ (which is again that of pointwise convergence), $G$ is metrizable.
Consider now two integers $k>0$ and $m\ge 0$, a uniformly continuous function $f\colon M^{k\times m} \to \mathbb{R}$, and $a ∈ M^k$. My question is, is the map \begin{align} Φ = Φ[f, a]\colon G &\to \mathbb{R}\\ T &\mapsto f(T^i a_j: i<m, j < k) \end{align} continuous with respect to $τ$?
My attempt
Here is my proof that $Φ$ is continuous. As $G$ is metrizable, it is enough to show that $Φ(T_n)$ converges to $Φ(T)$ in $\mathbb{R}$ for any sequence $T_n$ converging to $T$ in $τ$. The uniform continuity of $f$ may be regarded with respect to the max-distance $d_\infty$, so we just need to show that $(T_n^i a_j)_{ij}$ converges to $(T^i a_j)_{ij}$ in $d_\infty$. But now, for all $0<i<m$ and $j<k$, \begin{align} d(T_n^i a_j, T^i a_j) &\le d(T_n^i a_j, T_n T^{i-1} a_j) + d(T_n T^{i-1} a_j, T^i a_j)\\ &=d(T_n^{i-1} a_j, T^{i-1} a_j) + d(T_n (T^{i-1} a_j), T(T^{i-1} a_j) ) && \text{($T_n$ is isometric)}\\ &\le \sum_{\ell <i} d(T_n (T^{\ell} a_j), T(T^{\ell} a_j) ) =: s[i,j,n], && \text{(inductively)} \end{align} which goes to $0$ (as $n\to \infty$) because $T_n \to T$ in $τ$. It follows that $$ d_\infty\bigl( (T_n^i a_j)_{ij}, (T^i a_j)_{ij} \bigr) = \max_{i,j} d(T_n^i a_j, T^i a_j) \le \max_j s[m-1,j,n], $$ which goes to $0$ as $n\to \infty$, and this proves that $Φ$ is continuous.
Is my reasoning correct? Moreover, as $d$ is uniformly continuous, if we take $f\colon M^{2\times 2} \to \mathbb{R}$ defined by \begin{align} f \begin{pmatrix} x &y \\ x' & y' \end{pmatrix} = d(x', y), \end{align} and take $a ∈ M$ arbitrary and $a_T = T a $, then the continuity of $Φ[f,a,a_T]$ for all $a∈M$ in a topology $σ$ implies that any sequence $(T_n)_n \subseteq G$ converging to $T$ in $σ$, also converges pointwise to $T$. This would mean that the topology of pointwise convergence $τ$ on $G$ is equal to the initial topology with respect to the functions of the form of $Φ$.
