= Solution
A <vector bundle morphism> over $M$ is a smooth map $F:E\to E'$ with $\pi'\circ F=\pi$ whose restriction $E_p\to E'_p$ is linear for every $p$. A <vector bundle isomorphism> is such a map with a smooth bundle-morphism inverse. Equivalently, a fiberwise bijective smooth bundle morphism is an isomorphism: in local <vector bundle trivializations> it is multiplication by an invertible smooth <matrix>, and its inverse <matrix> is smooth.
For a <diffeomorphism> $f:N\to M$, define
$$
F:TN\longrightarrow f^*TM,\qquad v_p\longmapsto(p,df_p(v_p)).
$$
The derivative $df_p$ is a <linear isomorphism> by the <chain rule>, since $d(f^{-1})_{f(p)}\circ df_p=\operatorname{id}$. In <manifold charts>, $F$ is represented by the smooth <Jacobian matrix> of $f$, so it is a smooth bundle morphism. Its inverse is
$$
(p,w_{f(p)})\longmapsto d(f^{-1})_{f(p)}w_{f(p)},
$$
which is also smooth and fiberwise linear. Consequently
$$
\boxed{TN\cong f^*TM\quad\text{as vector bundles over }N.}
$$
This identifies the bundles over the same base $N$; the tangent map by itself is a map from $TN$ to $TM$ covering $f$.
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