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In a proof the professor did this while talking about the transformation matrix

$$ \Phi_{A'} \circ \tilde{T}_{A'}^{A} = id_V \circ \Phi_{A} = \Phi_{A} \rightarrow \quad \tilde{T}_{A'}^{A} = \Phi_{A'}^{-1} \circ \Phi_{A}$$

He basically just put $\Phi_{A'}$ on the other side, much like an algebraic transformation and said he was allowed to do it, because all the linear mappings above are isomorphisms.

Could somebody elaborate, why isomorphisms allow you do that?

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    Should that be $\tilde{T}_{A'}^A=\Phi_{A'}^{-1}\circ\Phi_A$? – 2017-01-31
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    Yes, you are correct, not only that, I think that answered my question. He just takes the inverse ( which he is allowed to do, because isomorphisms are bijective ) and the result follows. Is that correct? – 2017-02-01

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Since $\Phi_{A'}$ is an isomorphism, $id = \Phi^{-1}_{A'}\circ\Phi_{A'}$. Hence, $$\tilde{T}_{A'}^A = id\circ \tilde{T}_{A'}^A = \Phi^{-1}_{A'}\circ\Phi_{A'}\circ \tilde{T}_{A'}^A \overset{!}{=} \Phi^{-1}_{A'}\circ\Phi_A.$$ The equality (!) used the fact that $\Phi_{A'}\circ\tilde{T}_{A'}^A = \Phi_A$.

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    Sorry for the typo there, small but important mistake! Thank your for being so thorough – 2017-02-01
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    @JonathanZ Glad I could help! – 2017-02-01