On page 86 of Bishop and Goldberg's Tensor Analysis on Manifolds you are asked to show that contractions are invariants. Rather than doing it awkwardly for an (r,s)-tensor (r is the contravariant and s the covariant degree), I tried showing it for a (1,2)-tensor.
$e$ and $f$ are two different bases. $A^{f, i}_{jk}$ denotes the components of the tensor A with respect to the basis $f$.
Suppose $A^{f, i}_{jk} = b^i_ma^n_ja^p_kA^{e, m}_{np}$. Then $A^{f, \theta}_{\theta k} = b^\theta_m a^n_{\theta}a^p_kA^{e,m}_{np} = \delta^{n}_{m}a^p_kA^{e,m}_{np} = a^p_k A^{e,\theta}_{\theta p} = a^k_p A^{e,\theta}_{\theta k}$. Anything obviously wrong here? I'm expecting to get $ A^{f, \theta}_{\theta k}= A^{e,\theta}_{\theta k}$.
Actually you are done. You should not expect $ A^{f, \theta}_{\theta k}= A^{e,\theta}_{\theta k}$, the same way that you do not expect $A^{f, i}_{jk} = A^{e, i}_{jk}$. Indeed, saying that contraction is invariant mean
$$A^{f, \theta}_{\theta k}= a^j_kA^{e,\theta}_{\theta j}.$$
Indeed, writing $i, j, k, l, m, n,\cdots$ for both the indices of $e,f$ might confuses you. If you use, for example, $\alpha, \beta, \gamma, \delta, \epsilon, \cdots$ for indices of $e$, something like
$$ A^{f,i}_{i k}= A^{e,\alpha}_{\alpha k}$$
just don't make sense.