Does an automorphism map an endomorphism applied to a basis to the same endomorphism applied to another basis?

Question

A question inspired by Steven Roman's Advanced Linear Algebra:

Let $V$ be a vector space, $\sigma$ an automorphism on $V$ and $\tau$ an endomorphism on V.

We know $\sigma$ maps any basis of $V$ into a basis of $V$.

Now let $B =\{b_i\}$ be an ordered basis of $V$.

My question is: does there always exist an ordered basis of $V$, call it $C =\{c_i\}$, such that $\sigma$ maps $\tau(b_i)$ into $\tau(c_i)$ for each $i$?

Answer 1

This is not true for every order basis $B$ of $V$. Since $\tau$ is an endomorphism, there could be the case that $(\sigma \circ \tau)(b_i)$ is not in the image under $\tau$.