Given $H\leq G$ (finite) groups, and a (complex) representation $\theta:H\rightarrow GL(W)$ of $H$, we can define the induced representation $\rho:G\rightarrow GL(V)$ on $G$ by first choosing a coset representatives $\{r_1,\dots, r_n\}$ for $H$ in $G$, and then defining $$ V = \bigoplus_{i=1}^n r_iW. $$ For $g\in G$ and $1\leq i \leq n$, there exists a unique $1\leq j(i)\leq n$ and a unique $h_i\in H$ so that $gr_i = r_{j(i)} h_i$. Then $\rho_g$ is defined by extending the rule $\rho_g(r_iw) = r_{j(i)}\theta_{h_i}w$ to all of $V$ linearly.
I want to show that this construction is independent of choice of representatives.
Let $\{t_1, \dots, t_n\}$ be another choice of coset representatives for $H$ in $G$, and suppose $r_iH=t_iH$ for all $1\leq i\leq n$. Let $\{w_1, \dots, w_m\}$ be a basis for $W$. Then $\{r_iw_k\}$ is a basis for $V$, and $\{t_i w_k\}$ is a basis for $V' = \bigoplus_{i=1}^n t_iW.$
The map $r_iw_k\mapsto t_iw_k$ extends to a linear map $f:V\rightarrow V'$ which is clearly an isomorphism of vector spaces. But is $f$ an isomorphism of representations?
To that end, suppose $gt_i = t_{m(i)}h'_{i}$ so that $\rho'_g(t_i w_k) = t_{m(i)}\theta_{h'_i}$. Then $$\rho'_g f(r_iw_k) = \rho'_g(t_iw_k) = t_{m(i)}\theta_{h'_i}w_k$$ and $$f\rho_g(r_iw_k) = f(r_{j(i)}\theta_{h_i}w_k) = t_{j(i)}\theta_{h_i}w_k.$$
I've shown that $j(i)=m(i)$, so to show that $f$ is an isomorphism of representations we need to show that $\theta_{h_i} = \theta_{h'_i}$, i.e. that $h_i = h'_i$. But it doesn't seem that this always has to be true.
The statement $j(i) = m(i)$ is equivalent to the statement that $g$ permutes the summands of $V$ and $V'$ in the same way, i.e. that $\rho_g(r_i W) = r_{j(i)}W$ and $\rho'_g(t_i W) = t_{j(i)}W$, but it's unclear how this comes into play, if at all, in showing $V\cong V'$ as representations.
Is there better (natural) choice of map for $f$? Or is there a better way to show that this construction is independent of the choice of representations?