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I guess that isomorphic means diffeomorphic. Let me provide an example that seems conceptually different from the others given so far.

Take $M= \mathbb{R}^4 \times \mathbb{R}^4_{\operatorname{ex}} \times \mathbb{R}$, where $\mathbb{R}^4_{\operatorname{ex}}$ is any exotic $\mathbb{R}^4$.

You have two projections

$$ p_1 \colon M \longrightarrow \mathbb{R}^4, \quad p_2 \colon M \longrightarrow \mathbb{R}^4_{\operatorname{ex}},$$

whose fibres are diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}_{\operatorname{ex}}$ and $\mathbb{R}^4 \times \mathbb{R}$, respectively.

Since there exists no exotic $\mathbb{R}^5$ these fibres are both diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}$, but the bases are not diffeomorphic.

I guess that isomorphic means diffeomorphic. Let me provide an example that seems conceptually different from the others given so far.

Take $M= \mathbb{R}^4 \times \mathbb{R}^4_{\operatorname{ex}} \times \mathbb{R}$, where $\mathbb{R}^4_{\operatorname{ex}}$ is any exotic $\mathbb{R}^4$.

You have two projections

$$ p_1 \colon M \longrightarrow \mathbb{R}^4, \quad p_2 \colon M \longrightarrow \mathbb{R}^4_{\operatorname{ex}},$$

whose fibres are diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}_{\operatorname{ex}}$ and $\mathbb{R}^4 \times \mathbb{R}$, respectively.

Since there exists no exotic $\mathbb{R}^5$, these fibres are both diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}$, but the bases are not diffeomorphic.

I guess that isomorphic means diffeomorphic. Let me give an example that seems conceptually different from the others.

Take $M= \mathbb{R}^4 \times \mathbb{R}^4_{\operatorname{ex}} \times \mathbb{R}$, where $\mathbb{R}^4_{\operatorname{ex}}$ is any exotic $\mathbb{R}^4$.

You have two projections

$$ p_1 \colon M \longrightarrow \mathbb{R}^4, \quad p_2 \colon M \longrightarrow \mathbb{R}^4_{\operatorname{ex}},$$

whose fibres are diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}_{\operatorname{ex}}$ and $\mathbb{R}^4 \times \mathbb{R}$, respectively.

Since there exists no exotic $\mathbb{R}^5$ these fibres are both diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}$, but the bases are not diffeomorphic.

I guess that isomorphic means diffeomorphic. Let me provide an example that seems conceptually different from the others given so far.

Take $M= \mathbb{R}^4 \times \mathbb{R}^4_{\operatorname{ex}} \times \mathbb{R}$, where $\mathbb{R}^4_{\operatorname{ex}}$ is any exotic $\mathbb{R}^4$.

You have two projections

$$ p_1 \colon M \longrightarrow \mathbb{R}^4, \quad p_2 \colon M \longrightarrow \mathbb{R}^4_{\operatorname{ex}},$$

whose fibres are diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}_{\operatorname{ex}}$ and $\mathbb{R}^4 \times \mathbb{R}$, respectively.

Since there exists no exotic $\mathbb{R}^5$ these fibres are both diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}$, but the bases are not diffeomorphic.

I guess that isomorphic means diffeomorphic.

Take $M= \mathbb{R}^4 \times \mathbb{R}^4_{\operatorname{ex}} \times \mathbb{R}$, where $\mathbb{R}^4_{\operatorname{ex}}$ is any exotic $\mathbb{R}^4$.

You have two projections

$$ p_1 \colon M \longrightarrow \mathbb{R}^4, \quad p_2 \colon M \longrightarrow \mathbb{R}^4_{\operatorname{ex}},$$

whose fibres are diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}_{\operatorname{ex}}$ and $\mathbb{R}^4 \times \mathbb{R}$, respectively.

Since there exists no exotic $\mathbb{R}^5$ these fibres are both diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}$, but the bases are not diffeomorphic.

I guess that isomorphic means diffeomorphic. Let me give an example that seems conceptually different from the others.

Take $M= \mathbb{R}^4 \times \mathbb{R}^4_{\operatorname{ex}} \times \mathbb{R}$, where $\mathbb{R}^4_{\operatorname{ex}}$ is any exotic $\mathbb{R}^4$.

You have two projections

$$ p_1 \colon M \longrightarrow \mathbb{R}^4, \quad p_2 \colon M \longrightarrow \mathbb{R}^4_{\operatorname{ex}},$$

whose fibres are diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}_{\operatorname{ex}}$ and $\mathbb{R}^4 \times \mathbb{R}$, respectively.

Since there exists no exotic $\mathbb{R}^5$ these fibres are both diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}$, but the bases are not diffeomorphic.