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 $n \leq 3$.



Twisting bordism classes
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". 


