Twisting bordism classes

Question

Let $X$ be a reasonable topological space (I'd be happy to assume that $X$ is a smooth closed manifold) and let $f\colon M^n \rightarrow X$ be a continuous map from a smooth oriented $n$-manifold $M^n$ to $X$. Let $\phi\colon M^n \rightarrow M^n$ be an orientation-preserving diffeomorphism.

Question: Must it be the case that the maps $f\colon M^n \rightarrow X$ and $f \circ \phi \colon M^n \rightarrow X$ are oriented bordant to each other?

The reason I'm asking is that $f$ and $f \circ \phi$ induce the same element of $H_n(X;\mathbb{Z})$, and it is known that oriented bordism agrees with integral homology up to degree $3$.

If the answer to the question is negative, I'd also be interested in an explicit proof that this is the case for $n \leq 3$.

Answer 1

The correct definition of bordism should have this built in. More precisely, two singular manifolds $(M_0^n,f_0)$ and $(M_1^n,f_1)$ in a space $X$ are oriented bordant if there exists a smooth manifold $W^{n+1}$ with a diffeomorphism $\varphi: M_0\sqcup M_1\stackrel{\simeq}{\to}\partial W$ and a map $F:W\to X$ such that $f_i=(\partial F\circ \varphi)|_{M_i}$ for $i=0,1$.

It's true that some authors surpress this from the definition, but it's there for example in tom Dieck's recent book "Algebraic Topology".